Untuk menaikkan pH dari 100 ml larutan NH4Br 0,1 M

Berikut ini adalah pertanyaan dari human4932 pada mata pelajaran Kimia untuk jenjang Sekolah Menengah Atas

Untuk menaikkan pH dari 100 ml larutan NH4Br 0,1 M menjadi dua kali, ditambahkan larutan NH4OH 0,5 M (Kb = 10^-5) sebanyak…​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Penjelasan:

pH = 5, maka [H+] = 10-5

n CH3COOH = 0,1 M x 100 mL = 10 mmol

n NaOH = x mmol

begin mathsize 12px style space space space space space space space space space space space space space space space space space space space space space space space space space space CH subscript 3 COOH space space plus space space NaOH space space space space rightwards arrow space space space space CH subscript 3 COONa space space space plus space space space space space straight H subscript 2 straight O space space

Mula minus mula space space colon space space space 10 space mmol space space space space space space space space space space space straight x space mol space space space space space space space space space space space space space space space minus space space space space space space space space space space space space space space space space space space space space space space space space space space space minus space

Bereaksi space space space space space space space space colon space space minus straight x space mmol space space space space space space space space minus straight x space mmol space space space space space space space space space plus straight x space mmol space space space space space space space space plus straight x space mmol

Sisa space space space space space space space space space space space space space space space space colon space 10 minus straight x space mmol space space space space space space space space space minus space space space space space space space space space space space space space space space space space space straight x space mmol space space space space space space space space space space space straight x space mmol end style

[H+] = Ka.begin mathsize 12px style fraction numerator straight n straight space asam over denominator straight n straight space basa straight space konjugasi end fraction end style

[H+] = 10-5. begin mathsize 12px style fraction numerator 10 minus straight X over denominator straight X straight space end fraction end style

10-5 = 10-5. begin mathsize 12px style fraction numerator 10 minus straight X over denominator straight X straight space end fraction end style

begin mathsize 12px style 10 to the power of negative 5 end exponent over 10 to the power of negative 5 end exponent end style = begin mathsize 12px style fraction numerator 10 minus straight X over denominator straight X straight space end fraction end style

X = 5 mmol

[NaOH] = begin mathsize 12px style fraction numerator n space N a O H space over denominator V space N a O H space end fraction end style

V NaOH = begin mathsize 12px style fraction numerator n space N a O H space over denominator left square bracket N a O H right square bracket end fraction end style

V NaOH = begin mathsize 12px style fraction numerator 5 space m m o l over denominator 0 comma 05 space M space end fraction end style

V = 100 mL

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Last Update: Fri, 02 Jul 21