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Cover kimia kelas XI.cdr

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sehingga dengan adanya [OH - ] = 0,01 M :<br />

CH 3 COOH CH 3 COO– + H +<br />

mula-mula : 0,14M 0,06 M 0 M<br />

berubah : -x x x<br />

kesetimbangan : 0,14 – x 0,06 + x x<br />

[CH 3 COO - ] = 0,06 + x<br />

[CH 3 COOH] = 0,14 – x<br />

[H + ] = x M<br />

K<br />

a<br />

5<br />

(0, 06 x)( x)<br />

1,8 10 <br />

0,14 x<br />

(1,8 . 10-5)(0,14 - x) = x 2 + 0,06 x<br />

2,52 x 10 -6 - 1,8 x 10 -5 x = x 2 + 0,06 x<br />

x 2 + 0,06 x - 2,52 . 10 -6 = 0<br />

Pemecahan persamaan kuadratnya diperoleh :<br />

<br />

x <br />

2(1)<br />

2 6<br />

0,06 (0,06) 4(1)( 2,52 10 )<br />

x = 4,20 x 10 -5<br />

Sehingga : [H + ] = 4,20 x 10 -5 M<br />

pH = -log(4,20 x 10 -5 )<br />

pH = 4,38<br />

atau jika menggunakan persamaan :<br />

[ Garam basa ] <br />

pH log K a log <br />

[ Asam ]<br />

<br />

<br />

5 (0, 05 0, 01) <br />

pH log(1,8 10 ) log <br />

(0,15 0, 01)<br />

<br />

<br />

0, 06<br />

pH 4, 74 log <br />

0,14<br />

pH = 4,74 - 0,368 = 4,372<br />

<br />

Kesetimbangan Larutan 137

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