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Sebanyak 5 mL larutan  dengan pH 4 ditambah air sehingga pH nya menjadi 5. Volume larutan menjadi …. mL.

Pertanyaan

Sebanyak 5 mL larutan H C O O H left parenthesis K subscript a space equals space 4 space x space 10 to the power of negative sign 4 end exponent right parenthesis spacedengan pH 4 ditambah air sehingga pH nya menjadi 5. Volume larutan menjadi …. mL. space

Pembahasan Soal:

Langkah 1. Menghitung konsentrasi H C O O H awal (pH 4)

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 space end cell equals cell square root of K subscript a space x space open square brackets H C O O H close square brackets end root end cell row cell 10 to the power of negative sign 4 end exponent end cell equals cell square root of 4 space x space 10 to the power of negative sign 4 end exponent space x space open square brackets H C O O H close square brackets end root end cell row cell left parenthesis 10 to the power of negative sign 4 end exponent right parenthesis squared space end cell equals cell 4 space x space 10 to the power of negative sign 4 end exponent space x space open square brackets H C O O H close square brackets end cell row cell 10 to the power of negative sign 8 end exponent end cell equals cell 4 space x space 10 to the power of negative sign 4 end exponent space x space open square brackets H C O O H close square brackets end cell row cell open square brackets H C O O H close square brackets space end cell equals cell fraction numerator 10 to the power of negative sign 8 end exponent over denominator 4 space x space 10 to the power of negative sign 4 end exponent end fraction end cell row cell open square brackets H C O O H close square brackets space end cell equals cell space 2 comma 5 space x space 10 to the power of negative sign 5 end exponent space mol end cell row blank blank blank row blank blank blank end table

Langkah 2. Menghitung konsentrasi H C O O H (pH 5)

Error converting from MathML to accessible text.
 

Langkah 3. Menghitung volume akhir 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript bold 1 bold space italic x bold space V subscript bold 1 bold space end cell bold equals cell italic M subscript bold 2 bold space italic x bold space V subscript bold 2 end cell row cell 2 comma 5 space x space 10 to the power of negative sign 5 end exponent space mol space x space 5 space mL space end cell equals cell 2 comma 5 space x space 10 to the power of negative sign 7 end exponent space mol space x space V subscript 2 end cell row cell V subscript 2 space end cell equals cell fraction numerator 2 comma 5 space x space 10 to the power of negative sign 5 end exponent space mol space x space 5 space mL over denominator 2 comma 5 space x space 10 to the power of negative sign 7 end exponent space mol space end fraction end cell row cell V subscript 2 space end cell equals cell 500 space mL end cell row blank blank blank end table
 

Jadi, volume larutan akhir yang dihasilkan sebanyak 500 mL. space

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 19 Mei 2021

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