Didalam 500ml larutan terdapat 4,1 gram CH3COONa yang terlarut. Jika nilai Ka CH3COOH = 10 pangkat n.... Question from @wilinalim78

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October 2018 2 32 Report

Didalam 500ml larutan terdapat 4,1 gram CH3COONa yang terlarut. Jika nilai Ka CH3COOH = 10 pangkat negatif 5 ( pangkat -5 ) dan Ar Na=23, C=12, O=16, maka pH larutan tersebut adalah..

mohon bantuanya y..terima kasih

egayoza OH^- = √ (Kw ÷ Ka) × (gr ÷ Mr) × (1000 ÷ mL)
= √ (10^-14 ÷ 10^-5) × (4,1 ÷ 82) × (1000 ÷ 500)
= √ 10^-9 × 5 × 10^-2 × 2 = 10^-5
pOH = 5 ⇒ pH = 9

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$M_{CH_3COO^-}= \frac{gr/Mr}{V}= \frac{4,1/82}{0,5}=0,1~M \\ \\ ~[OH^-]= \sqrt{ \frac{Kw}{Ka}.[CH_3COO^-]} \\ \\ = \sqrt{ \frac{10^{-14}}{10^{-5}}.10^{-1}}= \sqrt{10^{-10}}=10^{-5} \\ \\ pOH=-log10^{-5}=5 \\ \\ pH=14-5=9$

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