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SMA Kimia: Pertanyaan, Jawaban, Latihan Soal & Pembahasan

Urutkan

Dalam 200 mL larutan asam propionat  terdapat  ion . Jika diketahui , massa asam propionat dalam larutan tersebut sebanyak...  

Pembahasan Soal:

Sebelum mencari massa asam propionat, maka harus mengetahui terlebih dahulu konsentrasi asam propionat. Konsentrasi asam propionat dapat diketahui melalui:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of Ka cross times M end root end cell row cell 7 cross times 10 to the power of negative sign 4 end exponent end cell equals cell square root of 1 comma 4 cross times 10 to the power of negative sign 5 end exponent cross times M end root end cell row cell open parentheses 7 cross times 10 to the power of negative sign 4 end exponent close parentheses squared end cell equals cell open parentheses square root of 1 comma 4 cross times 10 to the power of negative sign 5 end exponent cross times M end root close parentheses squared end cell row cell 49 cross times 10 to the power of negative sign 8 end exponent end cell equals cell 1 comma 4 cross times 10 to the power of negative sign 5 end exponent cross times M end cell row M equals cell begin inline style fraction numerator 49 cross times 10 to the power of negative sign 8 end exponent over denominator 1 comma 4 cross times 10 to the power of negative sign 5 end exponent end fraction end style end cell row M equals cell 3 comma 5 cross times 10 to the power of negative sign 2 end exponent space M end cell end table end style


Konsentrasi asam propionat, maka dapat digunakan untuk mengetahui massa asam propionat, melalui rumus konsentrasi:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row M equals cell begin inline style m over Mr end style cross times begin inline style fraction numerator 1000 over denominator V open parentheses mL close parentheses end fraction end style end cell row cell 3 comma 5 cross times 10 to the power of negative sign 2 end exponent end cell equals cell begin inline style m over 74 end style cross times begin inline style 1000 over 200 end style end cell row m equals cell begin inline style fraction numerator 3 comma 5 cross times 10 to the power of negative sign 2 end exponent cross times 74 cross times 200 over denominator 1000 end fraction end style end cell row blank equals cell 0 comma 518 space g end cell end table end style


Jadi, jawaban yang benar adalah B.

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