Pertanyaan

Bila 200 mL larutan NH 4 OH 0,2 M dicampurkan dengan 200 mL larutan NH 4 Cl 0,2 M dan K b NH 4 OH = 1 , 8 × 1 0 − 5 .Tentukan: pH larutan penyangga pH larutan penyangga setelah penambahan 10 mL HCI 0,1 M pH larutan penyangga setelah penambahan 10 mL NaOH 0,1 M

Bila 200 mL larutan $NH_{4} OH$ 0,2 M dicampurkan dengan 200 mL larutan $NH_{4} Cl$ 0,2 M dan $K_{b} NH_{4} OH = 1, 8 \times 1 0^{- 5}$ . Tentukan:

pH larutan penyangga
pH larutan penyangga setelah penambahan 10 mL HCI 0,1 M
pH larutan penyangga setelah penambahan 10 mL NaOH 0,1 M

I. Solichah

Master Teacher

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Pembahasan

pH larutan penyangga pH setelah penambahan 10 mL HCl 0,1 M pH setelah penambahan 10 mL NaOH0,1 M Dengan demikian, maka jawaban yang tepat adalah sesuai penjelasan di atas.

pH larutan penyangga
$n subscript N H subscript 3 end subscript equals 0 comma 2 cross times 200 equals 40 n subscript N H subscript 4 to the power of plus end subscript equals 0 comma 2 cross times 200 equals 40 open square brackets O H to the power of minus sign close square brackets equals K subscript b cross times fraction numerator open square brackets N H subscript 3 close square brackets over denominator open square brackets N H subscript 4 to the power of plus close square brackets end fraction open square brackets O H to the power of minus sign close square brackets equals 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator begin display style bevelled 40 over 400 end style over denominator bevelled 40 over 400 end fraction open square brackets O H to the power of minus sign close square brackets equals 1 comma 8 cross times 10 to the power of negative sign 5 end exponent pOH equals 5 minus sign log space 1 comma 8 pH equals 9 plus log space 1 comma 8$
pH setelah penambahan 10 mL HCl 0,1 M
$n subscript HCl equals 0 comma 1 cross times 10 equals 1 space mmol N H subscript 3 and H Cl yields N H subscript 4 Cl 40 space space space space space space space space 1 space space space space space space space space space space 40 bottom enclose negative sign 1 space space space space space space space minus sign 1 space space space space space space space plus 1 end enclose 39 space space space space space space space space space space space space space space space space space space space space space 41 open square brackets O H to the power of minus sign close square brackets equals 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator begin display style bevelled 39 over 410 end style over denominator begin display style bevelled 41 over 410 end style end fraction open square brackets O H to the power of minus sign close square brackets equals 1 comma 7 cross times 10 to the power of negative sign 5 end exponent pOH equals 5 minus sign log space 1 comma 7 pH equals 9 plus log space 1 comma 7$
pH setelah penambahan 10 mL NaOH 0,1 M
$n subscript NaOH equals 0 comma 1 cross times 10 equals 1 space mmol N H subscript 4 to the power of plus plus O H to the power of minus sign yields N H subscript 3 and H subscript 2 O 40 space space space space space space space space space 1 space space space space space space space space space space space 40 bottom enclose negative sign 1 space space space space space space space minus sign 1 space space space space space space space space space plus 1 end enclose 39 space space space space space space space space space space space space space space space space space space space space space space space 41 open square brackets O H to the power of minus sign close square brackets equals 1 comma 8 cross times 10 to the power of negative sign 5 end exponent cross times fraction numerator begin display style bevelled 41 over 410 end style over denominator begin display style bevelled 39 over 410 end style end fraction open square brackets O H to the power of minus sign close square brackets equals 1 comma 9 cross times 10 to the power of negative sign 5 end exponent pOH equals 5 minus sign log 1 comma 9 pH equals 9 plus log space 1 comma 9$

Dengan demikian, maka jawaban yang tepat adalah sesuai penjelasan di atas.