Roboguru

200 ml larutan   0,1 M dicampur dengan 50 ml larutan  0,1 M. Jika harga tetapan ionisasi asam (Ka) dan log 2 = 0,3 maka pH sebelum reaksi dan setelah reaksi berturut- turut adalah ...

Pertanyaan

200 ml larutan begin mathsize 14px style C H subscript 3 C O O H end style  0,1 M dicampur dengan 50 ml larutan begin mathsize 14px style open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca space end style 0,1 M. Jika harga tetapan ionisasi asam (Ka) dan log 2 = 0,3 maka pH sebelum reaksi dan setelah reaksi berturut- turut adalah ...

Pembahasan Soal:

menentukan pH asam lemah sebelum bereaksi :

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals square root of K italic a italic. italic M italic a end root open square brackets H to the power of plus sign close square brackets equals square root of left parenthesis 1 cross times 10 to the power of negative sign 5 end exponent right parenthesis cross times 0 comma 1 end root open square brackets H to the power of plus sign close square brackets equals square root of left parenthesis 1 cross times 10 to the power of negative sign 5 end exponent right parenthesis cross times 10 to the power of negative sign 1 end exponent end root open square brackets H to the power of plus sign close square brackets equals square root of 10 to the power of negative sign 6 end exponent right parenthesis end root open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 3 end exponent end style 

begin mathsize 14px style pH equals minus sign log open square brackets H to the power of plus sign close square brackets pH equals minus sign log left parenthesis 10 to the power of negative sign 3 end exponent right parenthesis pH equals 3 end style 

menentukan pH sesudah reaksi:

begin mathsize 14px style mol space C H subscript 3 C O O H space equals space M space cross times space V space equals space 0 comma 1 space cross times space 200 space equals space 20 space mmol space space mol space open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca space equals space M space cross times V space equals space 0 comma 1 space cross times 50 space equals space 5 space mmol end style 

begin mathsize 14px style open parentheses C H subscript 3 C O O close parentheses subscript 2 Ca yields 2 C H subscript 3 C O O to the power of minus sign and Ca to the power of 2 plus sign space left parenthesis valensi space garam space asam space equals 2 right parenthesis end style 

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals Ka fraction numerator mol space sisa space asam over denominator mol space garam space x space valen si end fraction open square brackets H to the power of plus sign close square brackets equals 1 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator 20 space mmol over denominator 5 mmol space x space 2 end fraction open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 5 end exponent cross times fraction numerator up diagonal strike 20 space mmol end strike over denominator up diagonal strike 10 space mmol end strike space end fraction open square brackets H to the power of plus sign close square brackets equals 2 cross times 10 to the power of negative sign 5 end exponent  pH space equals minus sign log open square brackets H to the power of plus sign close square brackets pH space equals minus sign log space 2 cross times 10 to the power of negative sign 5 end exponent pH space equals 5 minus sign log space 2 pH space equals 5 minus sign 0 comma 3 pH space equals 4 comma 7 space end style 

Jadi,  pH sebelum reaksi dan setelah reaksi berturut- turut adalah 3 dan 4,7.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 03 April 2021

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