Ke dalam 100 mL larutan dialirkan gas amonia hing...

Ke dalam 100 mL larutan $begin mathsize 14px style H subscript 2 S O subscript 4 space left parenthesis pH equals 0 comma 6990 right parenthesis end style$ dialirkan gas amonia hingga pH menjadi $begin mathsize 14px style 9 plus log space 2 end style$ . Berapa mL volume gas $begin mathsize 14px style N H subscript 3 end style$ yang dialirkan ke dalam larutan tersebut diukur pada $begin mathsize 14px style 0 degree C space dan space 76 space cmHg end style$ ? $begin mathsize 14px style left parenthesis K subscript b space N H subscript 3 equals 2 cross times 10 to the power of negative sign 5 end exponent right parenthesis end style$ $space$

Jawaban:

Ketika $begin mathsize 14px style H subscript 2 S O subscript 4 end style$ ditambahkan ke dalam gas amonia kemudian campuran tersebut memiliki pH $begin mathsize 14px style 9 plus log space 2 end style$ yang bersifat basa, berarti campuran tersebut adalah larutan penyangga yang berasal dari basa lemah dan garamnya. Amonia bereaksi dengan air menjadi $begin mathsize 14px style N H subscript 4 O H end style$ .

Mencari $begin mathsize 14px style italic M bold space H subscript bold 2 S O subscript bold 4 end style$

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell pH space H subscript 2 S O subscript 4 end cell equals cell 0 comma 6990 space end cell row cell pH space H subscript 2 S O subscript 4 end cell equals cell 1 minus sign log space 2 end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 2 cross times 10 to the power of negative sign 1 end exponent space M end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell italic a cross times M space H subscript 2 S O subscript 4 end cell row cell M space H subscript 2 S O subscript 4 end cell equals cell H to the power of plus over a end cell row cell M space H subscript 2 S O subscript 4 end cell equals cell fraction numerator 2 cross times 10 to the power of negative sign 1 end exponent M over denominator 2 end fraction end cell row cell M space H subscript 2 S O subscript 4 end cell equals cell 10 to the power of negative sign 1 end exponent space M end cell row blank blank blank end table end style$

Menghitung mol $begin mathsize 14px style H Cl bold space bold dan bold space bold mencari bold space O H to the power of bold minus sign end style$

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space H subscript 2 S O subscript 4 end cell equals cell V space H subscript 2 S O subscript 4 cross times M space H subscript 2 S O subscript 4 end cell row cell mol space H subscript 2 S O subscript 4 end cell equals cell 100 space mL cross times 0 comma 1 space M end cell row cell mol space H subscript 2 S O subscript 4 end cell equals cell 10 space mmol end cell row blank blank blank row cell pH space buffer end cell equals cell 9 plus log space 2 end cell row pOH equals cell 14 minus sign left parenthesis 9 plus log space 2 right parenthesis end cell row pOH equals cell 5 minus sign log space 2 end cell row cell O H to the power of minus sign end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent space M end cell end table end style$

Menghitung mol $begin mathsize 14px style N H subscript bold 3 end style$

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell O H to the power of minus sign end cell equals cell K subscript b cross times fraction numerator mol space N H subscript 4 O H over denominator mol space open parentheses N H subscript 4 close parentheses subscript 2 S O subscript 4 end fraction end cell row cell 2 cross times 10 to the power of negative sign 5 end exponent space M end cell equals cell 2 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator left parenthesis x minus sign 20 right parenthesis space mmol over denominator 10 space mmol end fraction end cell row cell 10 space mmol end cell equals cell x minus sign 20 space mmol end cell row x equals cell 10 space mmol plus 20 space mmol end cell row x equals cell 30 space mmol end cell row cell mol space N H subscript 3 end cell equals cell mol space N H subscript 4 O H end cell row cell mol space N H subscript 3 end cell equals cell 30 space mmol end cell end table end style$

Menghitung volume $begin mathsize 14px style N H subscript bold 3 end style$ dalam keadaan STP

$begin mathsize 14px style mol space N H subscript 4 O H double bond mol space N H subscript 3 mol space N H subscript 3 equals 30 space mmol V space N H subscript 3 double bond mol space N H subscript 3 cross times V subscript STP V space N H subscript 3 equals 0 comma 03 space mol cross times 22 comma 4 space L space mol to the power of negative sign 1 end exponent V space N H subscript 3 equals 0 comma 672 space L V space N H subscript 3 equals 672 space mL end style$

Jadi volume $begin mathsize 14px style N H subscript bold 3 end style$ yang dialirkan pada reaksi adalah 672 mL. $space$

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