Roboguru

Larutan asam klorida  0,1 M dan larutan  0,1 M dengan volume yang sama dicampur menjadi satu. Hitunglah pH campuran larutan tersebut!

Pertanyaan

Larutan asam klorida H Cl 0,1 M dan larutan H subscript 2 S O subscript 4 0,1 M dengan volume yang sama dicampur menjadi satu. Hitunglah pH campuran larutan tersebut!space 

Pembahasan Soal:

Kedua larutan tersebut merupakan senyawa asam karena pH less than 7. Perhitungan pH jenis ini tidak bisa direaksikan antara kedua spesi karena sama-sama memiliki ion proton open parentheses H to the power of plus sign close parentheses, pH campurannya dapat dihitung dengan cara berikut:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell Larutan space H Cl colon end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell a cross times M subscript HCl end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 1 cross times 0 comma 1 space M end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 0 comma 1 space M end cell row blank blank blank row blank blank cell Larutan space H subscript 2 S O subscript 4 colon end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell a cross times M subscript H subscript 2 S O subscript 4 end subscript end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 2 cross times 0 comma 1 space M end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 0 comma 2 space M end cell row blank blank blank row cell V subscript HCl end cell equals cell V subscript H subscript 2 S O subscript 4 end subscript double bond V end cell row cell M subscript a space total end subscript end cell equals cell fraction numerator n subscript total space dari space H to the power of plus sign over denominator V subscript total end fraction end cell row cell M subscript a space total end subscript end cell equals cell fraction numerator left parenthesis V cross times open square brackets H to the power of plus sign close square brackets subscript HCl right parenthesis plus left parenthesis V cross times open square brackets H to the power of plus sign close square brackets subscript H subscript 2 S O subscript 4 end subscript right parenthesis over denominator V and V end fraction end cell row cell M subscript a space total end subscript end cell equals cell fraction numerator left parenthesis V cross times 0 comma 1 space M right parenthesis plus left parenthesis V cross times 0 comma 2 space M right parenthesis over denominator 2 V end fraction end cell row cell M subscript a space total end subscript end cell equals cell fraction numerator 0 comma 1 V plus 0 comma 2 V over denominator 2 V end fraction end cell row cell M subscript a space total end subscript end cell equals cell fraction numerator 0 comma 3 V over denominator 2 V end fraction end cell row cell M subscript a space total end subscript end cell equals cell 0 comma 15 space M end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 1 comma 5 cross times 10 to the power of negative sign 1 end exponent space M end cell row blank blank blank row pH equals cell negative sign log space open square brackets H to the power of plus sign close square brackets end cell row pH equals cell negative sign log space open square brackets 1 comma 5 cross times 10 to the power of negative sign 1 end exponent space M close square brackets end cell row pH equals cell 1 minus sign log space 1 comma 5 end cell row blank blank blank end table   

Jadi pH campurannya adalah bold 1 bold minus sign bold log bold space bold 1 bold comma bold 5.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 04 Mei 2021

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