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Berapa gram  yang terlarut dalam 200 mL larutan  dengan pH = 4? Diketahui .

Pertanyaan

Berapa gram begin mathsize 14px style N H subscript 4 Cl end style yang terlarut dalam 200 mL larutan begin mathsize 14px style N H subscript 4 Cl end style dengan pH = 4? Diketahui begin mathsize 14px style K subscript italic b space N H subscript 3 equals 10 to the power of negative sign 5 end exponent end style.space 

Pembahasan Soal:

Senyawa begin mathsize 14px style N H subscript 4 Cl end style merupakan garam asam yang jika dilarutkan dalam air mengalami hidrolisis parsial dengan reaksi sebagai berikut.


begin mathsize 14px style N H subscript 4 Cl space rightwards arrow space N H subscript 4 to the power of plus space plus space Cl to the power of minus sign N H subscript 4 to the power of plus space plus space H subscript 2 O space rightwards harpoon over leftwards harpoon space N H subscript 4 O H space plus space H to the power of plus sign end style


Untuk menentukan massa larutan begin mathsize 14px style N H subscript 4 Cl end style yang terlarut dalam 200 mL larutan begin mathsize 14px style N H subscript 4 Cl end style dengan pH = 4 dapat dicari dengan langkah berikut.

  • Menentukan konsentrasi begin mathsize 14px style H to the power of plus sign end style dan molaritas begin mathsize 14px style N H subscript 4 Cl end style


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row pH equals cell 4 comma space maka end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 10 to the power of negative sign 4 end exponent end cell row blank blank blank row cell open square brackets H to the power of plus sign close square brackets end cell equals cell square root of Kw over Kb cross times M subscript kation end root end cell row cell 10 to the power of negative sign 4 end exponent end cell equals cell square root of 10 to the power of negative sign 14 end exponent over 10 to the power of negative sign 5 end exponent cross times M subscript kation end root end cell row cell 10 to the power of negative sign 8 end exponent end cell equals cell 10 to the power of negative sign 9 end exponent cross times M subscript kation end cell row cell M subscript kation end cell equals 10 row cell M subscript kation end cell equals cell M subscript N H subscript 4 to the power of plus end subscript double bond M subscript N H subscript 4 Cl end subscript end cell end table end style

 

  • Menentukan massa begin mathsize 14px style N H subscript 4 Cl end style yang terlarut


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M space N H subscript 4 Cl end cell equals cell fraction numerator massa space N H subscript 4 Cl over denominator Mr space N H subscript 4 Cl end fraction cross times 1000 over mL end cell row cell 10 space M end cell equals cell fraction numerator massa space N H subscript 4 Cl over denominator 53 comma 5 space g space mol to the power of negative sign 1 end exponent end fraction cross times fraction numerator 1000 over denominator 200 space mL end fraction end cell row cell massa space N H subscript 4 Cl end cell equals cell fraction numerator 10 space. space 53 comma 5 space.200 over denominator 1000 end fraction end cell row cell massa space N H subscript 4 Cl end cell equals cell 107 space gram end cell end table end style


Dengan demikian, masaa begin mathsize 14px style N H subscript 4 Cl end style yang terlarut dalam larutan tersebut adalah sebanyak 107 gram.space

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 15 Maret 2021

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