Pertanyaan

Sebanyak 0,316 gram ( CH 3 COO ) 2 Ca ( Mr = 158) dilarutkan dalam air hingga volumenya 1 liter. Jika Ka CH 3 COOH = 1 x 10 -5 , pH larutannya adalah ... .

Sebanyak 0,316 gram (CH3COO)2Ca (Mr = 158) dilarutkan dalam air hingga volumenya 1 liter. Jika Ka CH3COOH = 1 x  10-5, pH larutannya adalah ... .

  1. 6 - log 2

  2. 5 + log 2

  3. 8 - log 2

  4. 8 + log 2

  5. 9 + log 2

R. RGFREELANCETIGASATU

Master Teacher

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Pembahasan

pOH = - log[OH] = - log ( 2 x 10 -6 ) = 6 – log 2 pH = 14 – pOH = 14 – (6 – log 2) = 8 + log 2

M r space equals space fraction numerator g r over denominator M r end fraction cross times fraction numerator 1000 over denominator V space left parenthesis m L right parenthesis end fraction space equals space fraction numerator 0 comma 316 space g r over denominator 158 space g r divided by m o l end fraction cross times 1000 over 1000 space equals space 2 cross times 10 to the power of negative 3 end exponent space M  left square bracket O H to the power of minus right square bracket equals square root of fraction numerator K w over denominator K a end fraction cross times left square bracket G right square bracket cross times v a l e n s i end root space equals space square root of 10 to the power of negative 14 end exponent over 10 to the power of negative 5 end exponent open square brackets 2 cross times 10 to the power of negative 3 end exponent close square brackets cross times 2 end root space equals space 2 cross times 10 to the power of negative 6 end exponent

pOH = - log[OH] = - log ( 2 x 10-6) = 6 – log 2

pH = 14 – pOH = 14 – (6 – log 2) = 8 + log 2

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Hitunglah pH larutan CH 3 ​ COONa 0,008 M sebanyak 100 mL. Jika K a ​ CH 3 ​ COOH = 2 × 1 0 − 5 .

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