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Sebanyak 0,81 g HCN dan 0,64 g MgCN2​ dilarutkan dalam air hingga terbentuk 500 mL larutan. Ka​ HCN=2×10−4. (Ar H = 1, C = 12, N = 14, Mg = 24). Jika ke dalam larutan ditambahkan 1 mL HCl 0,005 M, berapakah pH larutan sekarang?

Pertanyaan

Sebanyak 0,81 g begin mathsize 14px style H C N end style dan 0,64 g begin mathsize 14px style Mg C N subscript 2 end style dilarutkan dalam air hingga terbentuk 500 mL larutan. begin mathsize 14px style K subscript a space H C N equals 2 cross times 10 to the power of negative sign 4 end exponent end style. (Ar H = 1, C = 12, N = 14, Mg = 24). Jika ke dalam larutan ditambahkan 1 mL HCl 0,005 M, berapakah pH larutan sekarang?

I. Solichah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban untuk pH larutan setelah penambahan HCl adalah 3,52.

Pembahasan

1. Menentukan mol begin mathsize 14px style H C N end style, begin mathsize 14px style Mg C N subscript bold 2 end style, dan begin mathsize 14px style C N to the power of bold minus sign end style. 

Persamaan umum:

begin mathsize 14px style n equals m over Mr end style  

  • mol begin mathsize 14px style H C N end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell fraction numerator 0 comma 81 space g over denominator Ar space H and Ar space C and Ar space N end fraction end cell row n equals cell fraction numerator 0 comma 81 space g over denominator left parenthesis 1 plus 12 plus 14 right parenthesis space g space mol to the power of negative sign 1 end exponent end fraction end cell row n equals cell fraction numerator 0 comma 81 space g over denominator 27 space g space mol to the power of negative sign 1 end exponent end fraction end cell row n equals cell 0 comma 03 space mol end cell end table end style 

  • mol begin mathsize 14px style Mg C N subscript 2 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell fraction numerator 0 comma 64 space g over denominator left parenthesis Ar space Mg and Ar space C plus 2 cross times Ar space N right parenthesis end fraction end cell row n equals cell fraction numerator 0 comma 64 space g over denominator left parenthesis 24 plus 12 plus 28 right parenthesis space g space mol to the power of negative sign 1 end exponent end fraction end cell row n equals cell fraction numerator 0 comma 64 space g over denominator 64 space g space mol to the power of negative sign 1 end exponent end fraction end cell row n equals cell 0 comma 01 space mol end cell end table end style 

  • mol undefined

undefined adalah basa konjugat dari begin mathsize 14px style H C N end style dan mol undefined diperoleh dari mol begin mathsize 14px style Mg C N subscript 2 end style berdasarkan reaksi berikut:

begin mathsize 14px style Mg C N subscript 2 equilibrium Mg to the power of 2 plus sign and 2 C N to the power of minus sign 0 comma 01 space space space space space space space space space 0 comma 01 space space space space space 0 comma 02 end style  

Jadi, mol undefined adalah 0,02 mol.

2. Menentukan mol HCl yang ditambahkan.

begin mathsize 14px style M equals n over V n double bond M middle dot V n equals 0 comma 005 space M middle dot 0 comma 001 space L n equals 0 comma 000005 space mol end style 

3. Menentukan mol begin mathsize 14px style H C N end style dan begin mathsize 14px style C N to the power of bold minus sign end style setelah penambahan HCl.

Apabila dalam larutan buffer ditambahkan HCl maka undefined akan bereaksi dengan begin mathsize 14px style H to the power of plus sign end style dari HCl, dan reaksi yang terjadi yaitu:

 

4. Menentukan nilai begin mathsize 14px style bold open square brackets H to the power of bold plus sign bold close square brackets end style setelah ditambahkan HCl.

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 end cell equals cell K subscript a middle dot fraction numerator n space asam over denominator n space basa space konjugat end fraction 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 4 end exponent middle dot fraction numerator 0 comma 030005 space mol over denominator 0 comma 019995 space mol end fraction end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 3 comma 00575 cross times 10 to the power of negative sign 4 end exponent end cell row cell open square brackets H to the power of plus sign close square brackets end cell equals cell 3 cross times 10 to the power of negative sign 4 end exponent end cell end table   

5. Menentukan nilai pH setelah ditambahkan HCl.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes 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 3 cross times 10 to the power of negative sign 4 end exponent end cell row pH equals cell 4 minus sign log space 3 end cell row pH equals cell 4 minus sign 0 comma 48 end cell row pH equals cell 3 comma 52 end cell end table end style 

Jadi, jawaban untuk pH larutan setelah penambahan HCl adalah 3,52.

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