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Berapa massa asam asetat CH 3 ​ COOH , jika Ka = 2 × 1 0 − 4 yang diperlukan untuk membuat 200 mL larutan asam asetat dengan pH = 5

Berapa massa asam asetat , jika Ka =  yang diperlukan untuk membuat 200 mL larutan asam asetat dengan pH = 5 

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N. Puspita

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massa yang dibutuhkan adalah .

massa begin mathsize 14px style C H subscript 3 C O O H end style yang dibutuhkan adalah begin mathsize 14px style 6 cross times 10 to the power of negative sign 6 end exponent space gram end style .

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Diketahui : V = 200 mL Ka = pH = 5 Ditanya : massa = ...? Penyelesaian soal : pH = 5 M = mol = M x Volume = Mr = 60 g/mol mol = massa = mol x Mr = = Jadi, massa yang dibutuhkan adalah .

Diketahui :

V begin mathsize 14px style C H subscript 3 C O O H end style = 200 mL

Ka = begin mathsize 14px style 2 cross times 10 to the power of negative sign 4 end exponent end style 

pH = 5

Ditanya : massa = ...?  

Penyelesaian soal :  

pH = 5

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 5 end exponent end style

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of Ka space x space M end root 10 to the power of negative sign 5 end exponent equals square root of 2 cross times 10 to the power of negative sign 4 end exponent space cross times space M end root open parentheses 10 to the power of negative sign 5 end exponent close parentheses squared equals 2 cross times 10 to the power of negative sign 4 end exponent cross times M 10 to the power of negative sign 10 end exponent equals space 2 cross times 10 to the power of negative sign 4 end exponent cross times M M space equals space fraction numerator 10 to the power of negative sign 10 end exponent over denominator 2 cross times 10 to the power of negative sign 4 end exponent end fraction M space equals space 0 comma 5 cross times 10 to the power of negative sign 6 end exponent end style     

M = begin mathsize 14px style mol over volume end style

mol = M x Volume = begin mathsize 14px style 0 comma 5 cross times 10 to the power of negative sign 6 end exponent space. space 0 comma 2 space L space equals space 0 comma 1.10 to the power of negative sign 6 end exponent space mol end style

Mr begin mathsize 14px style C H subscript 3 C O O H end style  = 60 g/mol

mol = begin mathsize 14px style massa over Mr end style

massa = mol x Mr = begin mathsize 14px style 0 comma 1 cross times 10 to the power of negative sign 6 end exponent cross times 60 space g forward slash mol end style = begin mathsize 14px style 6 cross times 10 to the power of negative sign 6 end exponent space gram end style 

Jadi, massa begin mathsize 14px style C H subscript 3 C O O H end style yang dibutuhkan adalah begin mathsize 14px style 6 cross times 10 to the power of negative sign 6 end exponent space gram end style .

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