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Perbandingan [HCO3−​] dengan [H2​CO3​] yang diperlukan untuk membentuk larutan penyangga dengan pH sebesar 7,7 adalah .... (Ka​ H2​CO3​=8×10−7)

Pertanyaan

Perbandingan begin mathsize 14px style left square bracket H C O subscript 3 to the power of minus sign right square bracket end style dengan begin mathsize 14px style open square brackets H subscript 2 C O subscript 3 close square brackets end style yang diperlukan untuk membentuk larutan penyangga dengan pH sebesar 7,7 adalah .... (begin mathsize 14px style K subscript a space H subscript 2 C O subscript 3 equals 8 cross times 10 to the power of negative sign 7 end exponent end style

  1. 1 : 10 undefined 

  2. 1 : 20 undefined 

  3. 1 : 40 undefined 

  4. 20 : 1 undefined 

  5. 40 : 1 undefined 

Y. Rochmawatie

Master Teacher

Jawaban terverifikasi

Pembahasan

Diketahui:

Akan dibuat larutan penyangga dengan pH sebesar 7,7

(begin mathsize 14px style K subscript a space H subscript 2 C O subscript 3 equals 8 cross times 10 to the power of negative sign 7 end exponent end style

Ditanya:

Perbandingan begin mathsize 14px style left square bracket H C O subscript 3 to the power of minus sign right square bracket end style dengan begin mathsize 14px style open square brackets H subscript 2 C O subscript 3 close square brackets end style yang diperlukan untuk membentuk larutan penyangga dengan pH sebesar 7,7 

Jawab:

1. Menghitung konsentrasi undefined larutan dengan pH=7,7

begin mathsize 14px style pH equals 7 comma 7 open square brackets H to the power of plus sign close square brackets equals 10 to the power of negative sign 7 comma 7 end exponent end style

2. Menghitung perbandingan begin mathsize 14px style left square bracket H C O subscript 3 to the power of minus sign right square bracket end style dan begin mathsize 14px style open square brackets H subscript 2 C O subscript 3 close square brackets end style dengan rumus penyangga asam

begin mathsize 14px style open square brackets H to the power of plus sign close square brackets equals K subscript a cross times fraction numerator open square brackets H subscript 2 C O subscript 3 close square brackets over denominator open square brackets H C O subscript 3 to the power of minus sign close square brackets end fraction 10 to the power of negative sign 7 comma 7 end exponent equals 8 cross times 10 to the power of negative sign 7 end exponent cross times fraction numerator open square brackets H subscript 2 C O subscript 3 close square brackets over denominator open square brackets H C O subscript 3 to the power of minus sign close square brackets end fraction fraction numerator 10 to the power of negative sign 7 comma 7 end exponent over denominator 8 cross times 10 to the power of negative sign 7 end exponent end fraction equals fraction numerator open square brackets H subscript 2 C O subscript 3 close square brackets over denominator open square brackets H C O subscript 3 to the power of minus sign close square brackets end fraction end style 

Agar perhitungan lebih mudah, kita balik perbandingan di atas menjadi:

begin mathsize 14px style fraction numerator open square brackets H C O subscript 3 to the power of minus sign close square brackets over denominator open square brackets H subscript 2 C O subscript 3 close square brackets end fraction equals fraction numerator 8 cross times 10 to the power of negative sign 7 end exponent over denominator 10 to the power of negative sign 7 comma 7 end exponent end fraction fraction numerator open square brackets H C O subscript 3 to the power of minus sign close square brackets over denominator open square brackets H subscript 2 C O subscript 3 close square brackets end fraction equals 8 cross times 10 to the power of 0 comma 7 end exponent end style 

Nilai dari begin mathsize 14px style 10 to the power of 0 comma 7 end exponent end style adalah 5,0119 atau kita bulatkan menjadi 5, sehingga:

begin mathsize 14px style fraction numerator open square brackets H C O subscript 3 to the power of minus sign close square brackets over denominator open square brackets H subscript 2 C O subscript 3 close square brackets end fraction equals 8 cross times 5 fraction numerator open square brackets H C O subscript 3 to the power of minus sign close square brackets over denominator open square brackets H subscript 2 C O subscript 3 close square brackets end fraction equals 40 over 1 end style 

Perbandingan begin mathsize 14px style left square bracket H C O subscript 3 to the power of minus sign right square bracket end style dengan begin mathsize 14px style open square brackets H subscript 2 C O subscript 3 close square brackets end style adalah 40:1

Jadi, jawaban yang tepat adalah E.

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