MODUL TEMA 9. Kese mbangan Ion dalam Larutan untuk Menentukan Kadar Suatu Zat

Modul Dinamis: Modul ini merupakan salah satu contoh bahan ajar pendidikan kesetaraan yang

berbasis pada kompetensi inti dan kompetensi dasar dan didesain sesuai kurikulum 2013. Sehingga modul ini merupakan dokumen yang bersifat dinamis dan terbuka lebar sesuai dengan kebutuhan dan kondisi daerah masing-masing, namun merujuk pada tercapainya standar kompetensi dasar.

Kimia Paket C - Setara SMA/MA kelas XI

Modul Tema 9 : Kesetimbangan Ion dalam Larutan untuk Menentukan Kadar Suatu Zat Penulis: Dra. Ernaviata,M.Pd

Diterbitkan oleh: Direktorat Pembinaan Pendidikan Keaksaraan dan

Ditjen Pendidikan Anak Usia Dini dan Pendidikan Masyarakat-Kementerian Pendidikan dan Kebudayaan, 2018

iv+ 86 hlm + illustrasi + foto; 21 x 28,5 cm

Hak Cipta © 2018 pada Kementerian Pendidikan dan Kebudayaan Dilindungi Undang-Undang

Kata Pengantar

P

endidikan kesetaraan sebagai pendidikan alternatif memberikan layanan kepada mayarakat yang

karena kondisi geografi s, sosial budaya, ekonomi dan psikologis tidak berkesempatan

mengiku-ti pendidikan dasar dan menengah di jalur pendidikan formal. Kurikulum pendidikan kesetaraan

dikembangkan mengacu pada kurikulum 2013 pendidikan dasar dan menengah hasil revisi berdasarkan

peraturan Mendikbud No.24 tahun 2016. Proses adaptasi kurikulum 2013 ke dalam kurikulum pendidikan

kesetaraan adalah melalui proses kontekstualisasi dan fungsionalisasi dari masing-masing kompetensi

dasar, sehingga peserta didik memahami makna dari setiap kompetensi yang dipelajari.

Pembelajaran pendidikan kesetaraan menggunakan prinsip fl exible learning sesuai dengan karakteristik

peserta didik kesetaraan. Penerapan prinsip pembelajaran tersebut menggunakan sistem pembelajaran

modular dimana peserta didik memiliki kebebasan dalam penyelesaian tiap modul yang di sajikan.

Kon-sekuensi dari sistem tersebut adalah perlunya disusun modul pembelajaran pendidikan kesetaraan yang

memungkinkan peserta didik untuk belajar dan melakukan evaluasi ketuntasan secara mandiri.

Tahun 2017 Direktorat Pembinaan Pendidikan Keaksaraan dan Kesetaraan, Direktorat Jendral Pendidikan

Anak Usia Dini dan Pendidikan Masyarakat mengembangkan modul pembelajaran pendidikan kesetaraan

dengan melibatkan Pusat Kurikulum dan Perbukuan Kemdikbud, para akademisi, pamong belajar, guru

dan tutor pendidikan kesetaraan. Modul pendidikan kesetaraan disediakan mulai paket A tingkat

kompe-tensi 2 (kelas 4 Paket A). Sedangkan untuk peserta didik Paket A usia sekolah, modul tingkat kompekompe-tensi 1

(Paket A setara SD kelas 1-3) menggunakan buku pelajaran Sekolah Dasar kelas 1-3, karena mereka masih

memerlukan banyak bimbingan guru/tutor dan belum bisa belajar secara mandiri.

Kami mengucapkan terimakasih atas partisipasi dari Pusat Kurikulum dan Perbukuan Kemdikbud, para

akademisi, pamong belajar, guru, tutor pendidikan kesetaraan dan semua pihak yang telah berpartisipasi

dalam penyusunan modul ini.

Jakarta, Desember 2018

Direktur Jenderal

Daftar Isi

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MODUL 9

Kesetimbangan Ion dalam Larutan

untuk

Menentukan Kadar suatu Zat

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Unit 9.1

Kesetimbangan Ion dalam Larutan Garam

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1DWULXP KLSRFORULW 1D2&O \DQJ WHUGDSDW GDODP ED\FOLQ GLJXQDNDQ VHEDJDL SHPXWLK

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1D&O 1D2+NXDW +&ONXDW PHUDK ELUX QHWUDO 1D+&2 1D2+NXDW +&2OHPDK ELUX ELUX EDVD !

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GLODUXWNDQ GDODP DLU PDND JDUDP WHUVHEXW DNDQ WHUXUDL PHQMDGL LRQLRQQ\D LRQLRQ GDULJDUDPLQLDNDQPHPSHQJDUXKLLRQ+_GDQLRQ2+ņ_{GDULDLUVHKLQJJDODUXWDQ\DQJ}

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\DQJ EHUHDNVL GHQJDQ DLU VHKLQJJD PHQ\HEDENDQ EHUWDPEDKQ\D LRQ 2+ņ GDODP

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&+&22ņ+2'&+&22+2+ņ

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EHUWDPEDK EHVDU GDQ PHQ\HEDENDQ S+ ODUXWDQ QDLN PHQMDGL ! ODUXWDQ EHUVLIDW EDVD*DUDP\DQJEHUDVDOGDULDVDPOHPDKGHQJDQEDVDNXDWWHUKLGUROLVLVVHEDJLDQ WHUKLGUROLVLVSDUVLDO

8QWXNUHDNVL&+&22ņ+2'&+&22+2+ņ

@ @> > @ @> > O H COO CH OH COOH CH K = ₋ − @ > @ @> > @ > ₋ − = COO CH OH COOH CH O H K @ > @ @> > − − = COO CH OH COOH CH Kh -LNDSHUVDPDDQ @ > @ @> > − − = COO CH OH COOH CH Kh GLNDOLNDQGHQJDQ @ > @ > + + H H _PDND @ @> > @ @> @> > + − + − = H COO CH H OH COOH CH Kh -LND_Kw >2+±@>+_@ @ > @ @> > COOH CH H COO CH Ka = − + 0DND xKw K Kh a = DWDX .HWHUDQJDQ Kw WHWDSDQNHVHWLPEDQJDQDLU ņ Ka WHWDSDQLRQLVDVLDVDPOHPDK

.RQVHQWUDVL &+&22+ VDPD GHQJDQ NRQVHQWUDVL 2+ņ NRQVHQWUDVL &+&22ņ

GLDQJJDSVDPDGHQJDQ NRQVHQWUDVL&+&22ņGDULLRQLVDVLJDUDP&+&22ņ\DQJ

WHUKLGUROLVLVGLDEDLNDQVHKLQJJDGDUL @ @> > @ @> > O H COO CH OH COOH CH K = ₋ − GLGDSDWNDQ a K Kw Kh₌ Lakmus merah berubah warna menjadi biru, larutan bersifat basa

@ > @ > @ > _{basa konjugasi} OH O H K ₌ − @ > @ > konjugasi basa OH Kh − = @ > @ >OH− ₌ K_h basakonjugasi _DWDX @ > @ > basa konjugasi K K OH a w = −

-LNDJDUDPQ\D%D&+&22PDNDEDVDNRQMXJDVLQ\DDGDODK>&+&22ņ@MDGLQ

PDND

'DSDWGLVLPSXONDQ7HWDSDQKLGUROLVLV.KGDQNRQVHQWUDVL2+ņ_DGDODK

8QWXNPRQRSURWLN 8QWXNSROLSURWLN MXPODKLRQOHELK GDULVDWX

6HEDQ\DN JUDP JDUDP QDWULXP VLDQLGD 1D&1 GLODUXWNDQ NH GDODP DLU KLQJJD YROXPHQ\D /LWHU -LND GLNHWDKXLKa +&1 ņ _{GDQAr 1D}

& 1 D 7XOLVNDQSHUVDPDDQUHDNVLKLGUROLVLV E +LWXQJKh F +LWXQJpHODUXWDQ Jawaban D &1ņ_DT+ 2l'+&1DT2+ņDT E a w K K Kh₌ − − − = = x x Kh F WDQ @ > laru mL x M gram NaCN r = @ >NaCN = x >1D&1@ 0 @ > @ > n basakonjugasi K K OH a w = − a

K

Kw

Kh

=

@ > @ > basa konjugasi K K OH a w = −

n = jumlah anion, Ba(CH3COO)2 n = 2

@

>

@

>

n

basa

konjugasi

K

K

OH

a w

=

−

_{> @ >}basakonjugasi _@ K K OH a w = − @ > x x x OH − ₌ −₋ >2+_{@ [} pOH log>2+@ pOH log[ pOH ±log pOH ± pH ± S+ -DGLS+ODUXWDQ1D&1DGDODK

6HEDQ\DN P/ ODUXWDQ .2+ 0 GLUHDNVLNDQ GHQJDQ P/ ODUXWDQ &+&22+0+LWXQJpHODUXWDQVHWHODKUHDNVLKa&+&22+

Jawaban

PPRO.2+ P/[0 PPRO PPRO&+&22+ P/[0 PPRO

.2+DT&+&22+DTĺ&+&22.DT+2l

PPROPPROPPRO

-DGLPPROJDUDPDWDX&+&22. PPRO

M mL mmol COOK CH @ > = = > @ >basakonjugasi @ K K OH a w = − @ > x OH − ₌ −₋ @ >OH − = x − >2+@ [

− p2+ log>2+ņ_@ p2+ log[ņ p2+ ±log p2+ ± -DGLp2+ pH ±pOH pH ± pH -DGLS+ODUXWDQVHWHODKUHDNVLDGDODK

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EDVD $J2+DT WHUKLGUROLVLV VHEDJLDQ GDODP DLU 3HUKDWLNDQ UHDNVL KLGUROLVLV GDUL $J&OEHULNXWLQL $J&ODT'$J_DT&Oņ_DT $J_DT+ 2l'$J2+DT+DT 5HDNVLWHUVHEXWPHQJKDVLONDQLRQ+_{VHKLQJJDODUXWDQEHUVLIDWDVDPGDQpH} 7HWDSDQKLGUROLVLV @ @> > @ @> > O H Ag H AgOH K = ₊ + @ > @ @> > @ > + + = Ag H AgOH O H K @ > @ @> > + + = Ag H AgOH Kh -LNDSHUVDPDDQ @ > @ @> > + + = AgOH_Ag H Kh GLNDOLNDQ @ > @ > − − OH OH _PDND @ @> > @ @> @> > − + − + = OH Ag OH H AgOH Kh -LND_Kw >+@>2+±@ @ > @ @> > AgOH OH Ag Kb = + − 0DND Kb Kw K x K Kh w b = =

.RQVHQWUDVL $J2+ VDPD GHQJDQ NRQVHQWUDVL LRQ + _{NRQVHQWUDVL LRQ $J}

GLDQJJDS VDPD GHQJDQ NRQVHQWUDVL LRQ $J _{GDUL LRQLVDVL JDUDP LRQ $J \DQJ}

WHUKLGUROLVLVGLDEDLNDQVHKLQJJD b K Kw Kh = Lakmus biru berubah warna menjadi merah, larutan bersifat asam 8QWXNOHELKPHPDKDPLPDWHULLQLFREDODKNHUMDNDQVRDOVHEDJDLEHULNXW

6HEDQ\DNJUDPJDUDP.DOVLXPDFHWDW&D&+&22GLODUXWNDQNHGDODP

DLU KLQJJD YROXPHQ\D /LWHU -LND GLNHWDKXL Ka &+&22+ ņ GDQ

Ar&D & 2 +

D 7XOLVNDQSHUVDPDDQUHDNVLKLGUROLVLV E +LWXQJKh

F +LWXQJpHODUXWDQ

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@ > @ > konjugasi asam H Kh ₌ + @ > @ >H+ = Kh asamkonjugasi $WDX_{> @} x_>asamkonjugasi _@ K K H b w = + 6HFDUDXPXP .HWHUDQJDQ Kw WHWDSDQNHVHWLPEDQJDQDLU ņ Kb WHWDSDQLRQLVDVLEDVDOHPDK Q MXPODKDVDPNRQMXJDVL

6HEDQ\DN JUDP DPPRQLXP NORULGD 1+&O GLODUXWNDQ NH GDODP DLU

VDPSDLYROXPHQ\DP/'LNHWDKXLKb1+DT ņGDQAr1 + &O D 7XOLVNDQSHUVDPDDQUHDNVLKLGUROLVLV E +LWXQJKh F +LWXQJpHODUXWDQ Jawab D 1+DT+2l'1+DT+2DT E b w K K Kh = − − = Kh _Kh F WDQ @ > _M x_mLlaru gram Cl NH r = NH Cl x M r @ > = = _{> @} x_>asam konjugasi _@ K K H b w = + @ > x H+ ₌ −₋ >H+@= x− >+@ [ņ pH log>+@ pH logņ pH ±log pH ± p+ -DGLS+ODUXWDQ1+&ODGDODK @ > @ > xn asam konjugasi K K H b w = +

6HEDQ\DNP/ODUXWDQ1+0GLFDPSXUGHQJDQP/ODUXWDQ+620 7HQWXNDQpHODUXWDQFDPSXUDQ 'LNHWDKXLKb1+DT [ņ Jawaban PPRO1+DT P/[0 PPRO PPRO+62DT P/[0 PPRO 1+DT+62DTĺ1+62DT PPROPPROPPRO -DGLPPRO1+62 PPRO M mL mmol SO NH = = 'DODP1+62DGDLRQ1+ _{> @} xn_>asamkonjugasi _@ K K H b w = + @ > x H+ ₌ −₋ >H+@= x− >+_@ pH log>+@ pH log pH ±log pH ± pH -DGLS+ODUXWDQ1+62DGDODK

*DUDPGDUL$VDP/HPDKGDQ%DVD/HPDK

*DUDP1+&1WHUEHQWXNGDUL+&1DVDPOHPDKGDQ1+DTEDVDOHPDK

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D +XEXQJDQ.KGHQJDQ.Z 5HDNVL1+&1'1+&1ņ 1++2'1+2++ &1ņ₊ 2'+&12+ņ 1+&1ņ+2'1+2++&1+2 1+&1ņ+2'1+2++&1 @ @> @> > @ @> > O H CN NH HCN OH NH Kh = _{+ −} GLNDOLNDQGHQJDQ @ @> > @ @> > − + − + OH H OH H @ @> @> > @ @> > O H CN NH HCN OH NH Kh= _{+ −} [ @ @> > @ @> > − + − + OH H OH H @ @> > @ > − + = OH NH OH NH Kh [ @ @> > @ > − + _CN H HCN [ @ > @ @> > O H OH H+ − E 3HQHQWXDQNRQVHQWUDVL+_GDQ2+ņ

3HUVDPDDQUHDNVL1+&1ņ+2'1+2++&1

@ @> > @ @> > − + = OH NH HCN OH NH Kh NDOLNDQ @ > @ > + + H H .DUHQD>+&1@ >1+2+@ >1+@ >&1ņ@ PDND @ @> > @ @> > − − = CN CN HCN HCN Kh [ @ > @ > + + H H @ > @ > @ > @ > + + − = x H H CN HCN Kh > @ _× + = H Ka Kh >+_{@ .K.D} @ > Ka Kb Ka Kw H+ = × Kb Ka Kw H @ > + ₌ DWDX +DUJD>+_@GDQ>2+ņ_{@EHUJDQWXQJSDGDKDUJD.DGDQ.E}

/DUXWDQ JDUDP \DQJ EHUDVDO GDUL DVDP OHPDK GDQ EDVD OHPDK GDSDW EHUVLIDW DVDPEDVDDWDXQHWUDObergantung pada harga Ka dari asam dan Kb dari basa Ka Kb Kw Kh =

Ka Kb Kw H+_@= × >

>OHB@= Kw_Ka×Kb

7HQWXNDQpH ODUXWDQ 1+&+&22 0 MLND

K

a&+&22+ ņ GDQ b

K

1+ ņ Jawaban b w a K K K H+_@= > @ > ₋ − − + ₌ x H >+@ pH log>+@ pH log pH

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'DODPNHKLGXSDQVHKDULKDULSHULVWLZDKLGUROLVLVJDUDPEDQ\DNPDQIDDWQ\D

%HULNXW EHEHUDSD KDO \DQJ PXQJNLQ SHUOX NLWD NHWDKXL SHQHUDSDQ KLGUROLVLV JDUDPGDODPNHKLGXSDQVHKDULKDUL

'DODP WXEXK NLWD WHUMDGL UHDNVL KLGUROLVLV 3URVHV SHQFHUQDDQ PDNDQDQ GDODP WXEXK PDQXVLD GDQ KHZDQ PHUXSDNDQ UHDNVL KLGUROLVLV &RQWRK DPLOXP WHUKLGUROLVLV PHQMDGL JOXNRVD SURWHLQ WHUKLGUROLVLV PHQMDGL DVDP DPLQR OHPDNWHUKLGUROLVLVPHQMDGLDVDPOHPDNGDODPXVXVKDOXV

Gambar 9.3 Usus halus Sumber: www slideshare.net

3HODUXWDQ 6DEXQ JDUDP QDWULXP VWHDUDW &+&221D

VDEXQ FXFL DNDQ PHQJDODPL KLGUROLVLV MLND GLODUXWNDQ GDODP DLU PHQJKDVLONDQDVDPVWHDUDWGDQEDVDQ\D1D2+

5HDNVL&+&221D+2'&+&22+1D2+

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Gambar 9.4 Sabun

Sumber: www slideshare.net

3HQMHUQLKDQ DLU PLQXP ROHK SHUXVDKDDQ DLU PLQXP 3$0 EHUGDVDUNDQ SULQVLS KLGUROLVLV \DLWX PHQJJXQDNDQ VHQ\DZD DOXPLQLXP IRVIDW \DQJ PHQJDODPL KLGUROLVLVWRWDO

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5HDNVL\DQJWHUMDGL$O_{DT+2l'$O2+V+DT}

3XSXNGLEHULNDQ SDGD WDQDPDQ DJDU WXPEXK VXEXU PDNDS+WDQDKKDUXVGLMDJDGDQGLVHVXDLNDQGHQJDQS+ WDQDPDQ 2OHK NDUHQD LWX GLSHUOXNDQ SXSXN \DQJ GDSDW PHQMDJD S+ WDQDK DJDU WLGDN WHUODOX DVDP DWDX EDVD VHSHUWL SXSXN XUHD 763 =$ %LDVDQ\D SDUD SHWDQL PHQJJXQDNDQ SHOHW SDGDW 1+62XQWXN PHQXUXQNDQ

S+WDQDK*DUDP1+62EHUVLIDWDVDPLRQ1+DNDQ

WHUKLGUROLVLV GDODP WDQDK PHPEHQWXN 1+GDQ +\DQJ

EHUVLIDWDVDP

Gambar 9.5: pupuk

Sumber: www.coolboxdelta.com

3HPXWLK 3DNDLDQ NLWD MXJD VHULQJ PHPDNDL SHPXWLK SDNDLDQ 3URGXN LQL PHQJDQGXQJ NLUDNLUD 1D2&O \DQJ VDQJDW UHDNWLI VHKLQJJD GDSDW PHQJKDQFXUNDQ SHZDUQD VHKLQJJD SDNDLDQ PHQMDGL SXWLKNHPEDOL*DUDPLQLWHUEHQWXNGDULDVDPOHPDK+2&OGHQJDQEDVD NXDW 1D2+ ,RQ 2&Oņ_{WHUKLGUROLVLV PHQMDGL +2&O GDQ 2+}ņ _VHKLQJJD

JDUDP1D2&OEHUVLIDWEDVD Gambar 9.6: bayclin Sumber: www slideshare.net

Menganalisis Sifat Larutan Garam dengan Menggunakan Lakmus

dan Indikator Universal

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Penugasan 1

Tujuan

Teori

6HGLDNDQODK PLQLPDO PDFDP ODUXWDQ \DQJ DQGD WHPXNDQ GDODP NHKLGXSDQ VHKDULKDUL\DQJPHQJDQGXQJJDUDPVHSHUWLULQVRPHQJDQGXQJJDUDP1D32 JDUDPGDSXU1D&OGDQcarilah larutan lain\DQJDNDQGLXMLGDODPWDEXQJUHDNVL &HOXSNDQNHUWDVODNPXVPHUDKGDQELUXNHGDODPODUXWDQOLKDWSHUXEDKDQZDUQD

NHUWDVODNPXV&DWDWSDGDOHPEDUSHQJDPDWDQ

&HOXSNDQ NHUWDV LQGLNDWRU XQLYHUVDO NHGDODP ODUXWDQ NHPXGLDQ FRFRNNDQ SHUXEDKDQ ZDUQD NHUWDV LQGLNDWRU GDQ FRFRNNDQ GHQJDQ ZDUQD \DQJ DGD SDGD VDPSXOLQGLNDWRUXQLYHUVDO&DWDWS+VHVXDLGHQJDQZDUQDWHUVHEXW

Alat dan bahan

$ODWWDEXQJUHDNVLUDNWDEXQJUHDNVLNHUWDVODNPXVLQGLNDWRU8QLYHUVDO %DKDQEHUEDJDLPDFDPODUXWDQ\DQJPHQJDQGXQJJDUDP

7DEHOSHQJDPDWDQ /DUXWDQ *DUDP\DQJ WHUNDQGXQJ GDODPODUXWDQ :DUQD ODUXWDQ 3HUXEDKDQZDUQDGDODPLQGLNDWRU 6LIDW /DUXWDQ $VDP%DVD $QDOLVLV'DWD /DNPXV PHQMDGL ,QGLNDWRU8QLYHUVDO SDGD$QJND 5LQVR 1D32 %HQLQJ %LUX %DVD 1D32EHUDVDOGDUL1D2+ EDVDNXDWGDQ+32DVDP OHPDKMDGLWHUKLGUROLVLV SDUVLDOEHUVLIDWEDVD WHUEXNWLS+ « « « « « « « « « Langkah-Langkah

Catatan

bila anda tidak mempunyaikertas lakmus atau indikator universal, apa yang dapat anda lakukan? Ya Anda masih ingatkan dengan indikator alam? (Modul 8 tentang indikator) Ya ….anda dapat memanfaatkan kunyit, bunga kembang sepatu, atau indikator alam lainnya untuk mengidentifikasi garam-garam yang akan anda uji. Untuk pengganti Indikator Universal anda dapat menggunakan ekstrak kol merah

5XEULN3HQLODLDQ

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=

=

=

diperoleh

x

x

maksimum

yang

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'ĂŵďĂƌϵ͘ϳ͗dekstur tanah Air kapur Tepi pantai yang indah

Sumber: www.tneutron .net perpustakaancyber.blogspot.co.id. artikelmateri.blogspot.com

$SD \DQJ GDSDW DQGD DPDWL GDUL JDPEDU WHUVHEXW" $SD \DQJ DQGD SLNLUNDQ WHQWDQJ WHNVWXU WDQDK DLU NDSXU GDQ SHQJHQGDSDQ GL WHSL SDQWDL" 6HFDUD VHGHUKDQD VDQJDW PHQDULNNDQ" 6HWHODK PHPEDKDV PDWHUL WHQWDQJ NHODUXWDQ GDQ KDVLONDOL NHODUXWDQ LQL WHQWXQ\D DQGD DNDQ GDSDW PHQMDZDESHUWDQ\DDQLQLGHQJDQEDLN

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Larutan jenuh \DQJ PHQJDQGXQJ NULVWDO ]DW SDGDW WLGDN ODUXW WHUGDSDW GDODPkeadaan

kesetimbangan dinamik GHQJDQ PROHNXOPROHNXO DWDX LRQLRQQ\D GDODP ODUXWDQ .HODUXWDQ

GLSHQJDUXKLoleh jenis zat terlarut, jenis pelarut dan suhu.

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Anda pasti senang melihat pantai ini kan ? bagaimanakah proses terbentuknya pantai yang indah ini? Sedimentasi ini terjadi melalui proses pengendapan material yang ditransport oleh media air, angin, di suatu cekungan. Delta yang terdapat di mulut-mulut sungai adalah hasil dan proses pengendapan material-material yang diangkut oleh air sungai.

Carilah informasi lebih lanjut, buatlah makalah sederhana dan presentasikan dalam kelompok belajarmu

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−@sp = solubility product

Ksp = hasil kali kelarutan (solubility product constant)

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Ksp GLGHILQLVLNDQ VHEDJDL KDVLO NDOL NRQVHQWUDVL LRQLRQ VXDWX HOHNWUROLW GDODP ODUXWDQ \DQJ

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WHUODUXWGDODPDLUMakin besar harga Ksp, maka zat

elektrolit tersebut makin mudah larut. +XEXQJDQ.HODUXWDQ'HQJDQKsp

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PHODUXW

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MHQXK

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PHQJHQGDS

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GLWXUXQNDQVHEDJDLEHULNXW0LVDONDQNHODUXWDQ$J&OGDODPDLU VPRO/

Ksp = [A

Y+

_]

X

_[

_B

X−

_]

Y

cobalah Andaanalisis bagaimana kelarutan dari kapur (CaCO3) dan gips (CaSO4). Ambil zat tersebut dalam jumlah yang sama dan larutkan masing-masing dalam air yang sama pula, kemudian diaduk, diamkan. Manakah yang lebih mudah larut? Manakah yang mempunyai Ksp yang lebih besar? (lihat tabel 9.1 benarkah analisis anda?

PDND

$J&OV'$J_DT&O−_DT

VPRO

/ VPRO/ VPRO/

Ksp$J&O >$J@>&O−@

Ksp$J&O VPRO/[VPRO/

Ksp$J&O VPRO/

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$J32V'$JDT32−DT VVV .VS >$J@>32−@ − _VV − _V V − V −

NHODUXWDQ$J32VDGDODKPRO/LWHU

-DGLNHODUXWDQ$J32VGDODPP/DGDODK/LWHU[PRO/LWHU −PRO

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(

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laru v x Mr gram M= −₌ gram_x

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%DJDLPDQD KXEXQJDQ DQWDUD WHWDSDQ KDVLO NDOL NHODUXWDQ Ksp GHQJDQ UHDNVL

SHQJHQGDSDQ"3HUKDWLNDQFRQWRKEHULNXW

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VHKLQJJD DUDK NHVHWLPEDQJDQ DNDQ EHUJHVHU NH DUDK$J&ODNLEDWQ\D_{NHODUXWDQsemakin kecil}

%DQGLQJNDQNHODUXWDQ$J&OGDODPDLUPXUQLGDQNHODUXWDQ$J&OGDODP1D&O0MLND GLNHWDKXLKsp$J&O [− Jawaban D .HODUXWDQ$J&OGDODPDLUPXUQL $J&O'$J_DT&O−_DT VV .VS >$J@>&O−@ V[V [− _V s= x− V [− -DGLNHODUXWDQ$J&OGDODPDLUPXUQL [−₀ E .HODUXWDQ$J&OGDODPODUXWDQ1D&O0 'DODPODUXWDQ1D&OWHUMDGLUHDNVLEHULNXW

Coba anda ambil satu sendok kapur (CaCO3) larutkan dalam setengah gelas air dan diaduk, diamkan. Apakah mengendap? kemudian tandai tinggi endapan.

Anda ambil satu sendok soda kue (NaHCO3) dan larutkan dalam air kemudian diaduk, tuangkan kedalam larutan kapur tadi, dan aduk lagi, diamkan, bagaimana dengan endapan kapur tadi? Diskusikan dalam kelompok anda? 8QWXNOHELKPHPDKDPLNRQVHSSHQJHQGDSDQFREDODKNHUMDNDQVRDOEHULNXW $SDNDKWHUEHQWXNHQGDSDQ D MLNDNHGDODPP/1D&O−_{0GLWDPEDKNDQP/ODUXWDQ$J12} −0.VS $J&O− E MLNDNHGDODPP/3E120GLWDPEDKNDQP/ODUXWDQ1D&O0.VS 3E&O− .VS0J2+−MLNDSDGDODUXWDQ0J&O0GLWDPEDKNDQODUXWDQ1D2+%HUDSD S+SDGDVDDWPXODLWHUEHQWXNHQGDSDQ"

1D&OVĺ1D_DT&O−_DT

000

VVV

%HUDUWL GDODP ODUXWDQ WHUGDSDW LRQ &O− \DQJ EHUDVDO GDUL 1D&O 0 GDQ $J&O .RQVHQWUDVL LRQ &O− GDUL $J&O \DLWX [ − _{GDSDW GLDEDLNDQ NDUHQD VDQJDW NHFLO}

GLEDQGLQJNDQGHQJDQNRQVHQWUDVLLRQ&O−GDUL1D&O\DLWX0 Ksp >$J@>&O−@ [− _>$J@[0 Ag x M @ > − − + _{= =}

.HODUXWDQ $J&O GDODP 1D&O 0 [ − _{0 OHELK NHFLO GDUL NHODUXWDQ $J&O}

GDODPDLUPXUQL\DLWX[−_{0-DGLWHUEXNWLEDKZDLRQVHQDPD\DLWX&O}−_PHPSHUNHFLO

NHODUXWDQJDUDP\DQJVXNDUODUXW

Tabel 9.1 Daftar tabel Hasil Kali Kelarutan Endapan –endapan Pada suhu Kamar

=DW .VS =DW .VS =DW .VS =DW .VS

$J%U ņ _&D&2

4,8.10 ɔ9 )H6 4,0.10 ɔ19 3E&O 2,4.10 ɔ4

$J%U2 ņ &D&2 2,6.10 ɔ9 +J%U 5,2.10 ɔ23 3E&2 3,3.10 ɔ14

$J&16 ņ _&D)

3,2.10 ɔ11 +J&O 3,5.10 ɔ18 3E&U2 1,8.10 ɔ14

$J&O ņ _&D62

2,3.10 ɔ4 +J, 1,2.10 ɔ28 3E) 3,7.10 ɔ8

$J&2 ņ &G6 1,4.10 ɔ28 +J6 1,0.10 ɔ45 3E, 8,7.10 ɔ 9

$J&U2 ņ &R2+ 1,6.10 ɔ18 +J6 4,0.10 ɔ54 3E32 1,5.10 ɔ32

$J, ņ _&R2+

2,5.10 ɔ43 .3W&O 1,1.10 ɔ5 3E6 5,0.10 ɔ29

$J,2 ņ &R6 3,0.10 ɔ 26 0J&2 1,0.10 ɔ 5 3E62 2,2.10 ɔ8

$J32 ņ &U2+ 2,9.10 ɔ29 0J&2 8,6.10 ɔ 5 6U&2 1,6.10 ɔ 9

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Sumber: VOGEL, Buku teks Analisis Anorganik Kualitatif Makro dan semimikro, edisi ke lima

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