Pertanyaan

Untuk membuat larutan penyangga dengan pH = 9, maka ke dalam 400 mL larutan NH 3 0,5 M (Kb = 10 -5 ) harus ditambah larutan HCl 1 M sebanyak....

Untuk membuat larutan penyangga dengan pH = 9, maka ke dalam 400 mL larutan NH₃ 0,5 M (Kb = 10^-5) harus ditambah larutan HCl 1 M sebanyak....

100 mL
200 mL
300 mL
400 mL
500 mL

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J. Siregar

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

Jawaban terverifikasi

Pembahasan

pH = 9 pOH = 14 – 9 pOH = 5 [OH - ] = 10 -5 mol NH 3 = M X V = 0,5 M x 400 mL = 200 mmol = 0,2 mol Konsentrasi HCl = 1 M Mol HCl= x mol NH 3 + HCl NH 4 Cl Mula – mula : 0,2 x Reaksi : -x -x +x Setimbang : 0,2-x - x mol HCl = 0,1 mol volume HCl = mol konsentrasi = 0,1 Liter = 100 mL

pH = 9

pOH = 14 – 9

pOH = 5

[OH^-] = 10^-5

mol NH₃ = M X V

= 0,5 M x 400 mL = 200 mmol = 0,2 mol

Konsentrasi HCl = 1 M

Mol HCl = x mol

NH₃ + HCl NH₄Cl

Mula – mula : 0,2 x

Reaksi : -x -x +x

Setimbang : 0,2-x - x

$table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets O H to the power of minus close square brackets end cell equals cell K b cross times fraction numerator m o l space b a s a over denominator m o l space g a r a m end fraction end cell row cell 10 to the power of negative 5 end exponent space space space end cell equals cell 10 to the power of negative 5 end exponent cross times fraction numerator 0 comma 2 minus x over denominator x end fraction end cell row cell 1 space space space space space space space space space end cell equals cell fraction numerator 0 comma 2 minus x over denominator x end fraction end cell row cell x space space space space space space space space space end cell equals cell 0 comma 2 minus x end cell row cell 2 x space space space space space space space end cell equals cell 0 comma 2 end cell row cell x space space space space space space space space space end cell equals cell 0 comma 1 end cell end table$

mol HCl = 0,1 mol

volume HCl = $fraction numerator m o l over denominator k o n s e n t r a s i end fraction equals fraction numerator 0 comma 1 space m o l over denominator 1 space M end fraction$ = 0,1 Liter = 100 mL