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Tentukan massa  yang bereaksi dengan 200 mL larutan  0,2 M. Jika pH campurannya = 9  dan Ar N=14, Ar O=16, dan Ar H=1.

Pertanyaan

Tentukan massa N H subscript 4 O H yang bereaksi dengan 200 mL larutan N H subscript 4 Cl 0,2 M. Jika pH campurannya = 9 left parenthesis Kb space N H subscript 4 O H equals 10 to the power of negative sign 5 end exponent right parenthesis dan Ar N=14, Ar O=16, dan Ar H=1. space 

Pembahasan Soal:

Campuran antara basa lemah (N H subscript 4 O H) dengan garamnya (N H subscript 4 Cl) merupakan larutan penyangga basa. Massa N H subscript 4 O H yang dibutuhkan dapat ditentukan dengan cara sebagai berikut.

1. Menentukan konsentrasi ion O H to the power of minus sign


table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 9 comma space maka colon end cell row pOH equals cell 14 minus sign pH end cell row blank equals cell 14 minus sign 9 end cell row blank equals 5 row blank blank blank row pOH equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row 5 equals cell negative sign log open square brackets O H to the power of minus sign close square brackets end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell 10 to the power of negative sign 5 end exponent end cell end table


2. Menentukan mol N H subscript 4 O H


table attributes columnalign right center left columnspacing 0px end attributes row cell n space N H subscript 4 Cl end cell equals cell M cross times V end cell row blank equals cell 0 comma 2 space M cross times 200 space mL end cell row blank equals cell 40 space mmol end cell row blank blank blank row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell K subscript b cross times fraction numerator n space basa over denominator n space garam middle dot valensi end fraction end cell row blank equals cell K subscript b cross times fraction numerator n space N H subscript 4 O H over denominator n space N H subscript 4 Cl middle dot 1 end fraction end cell row cell 10 to the power of negative sign 5 end exponent end cell equals cell 10 to the power of negative sign 5 end exponent cross times fraction numerator n space N H subscript 4 O H over denominator 40 space mmol end fraction end cell row cell n space N H subscript 4 O H end cell equals cell 40 space mmol end cell end table


3. Menentukan massa N H subscript 4 O H


table attributes columnalign right center left columnspacing 0px end attributes row cell Mr space N H subscript 4 O H end cell equals cell 1 point Ar space N and 5 point Ar space H and 1 point Ar space O end cell row blank equals cell 1.14 plus 5.1 plus 1.16 end cell row blank equals cell 14 plus 5 plus 16 end cell row blank equals cell 35 space bevelled g over mol end cell row blank blank blank row cell n space N H subscript 4 O H end cell equals cell 40 space mmol equals 0 comma 04 space mol end cell row cell massa space N H subscript 4 O H end cell equals cell left parenthesis n cross times Mr right parenthesis N H subscript 4 O H end cell row blank equals cell 0 comma 04 space mol cross times 35 space bevelled g over mol end cell row blank equals cell 1 comma 4 space g end cell end table


Jadi, massa N H subscript 4 O H yang bereaksi dengan 200 mL larutan N H subscript 4 Cl 0,2 M. Jika pH campurannya = 9 adalah 1,4 gram.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 05 Mei 2021

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