9. Larutan CH3COOH 0,1 M, sebanyak 75 mL dicampur dengan 50 mL larutan NaOH 0,1

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M. Jika diketahui Ka asam asetat 2.10-5, maka pH larutan yang terbentuk
adalah….
A. 5- log 2
B.5
C.5+ log2
D. 9
E. 9 + log​

9. Larutan CH3COOH 0,1 M, sebanyak 75 mL dicampur dengan 50 mL larutan NaOH 0,1

Jawaban:

Mmol CH3COOH= 75×0.1 = 7.5

Mmol NaOH. = 50×0.1 = 5

CH3COOH + NaOH —-> CH3COONa + OH-

m 7.5. 5

r. 5. 5. 5. –

———————————————————–

2.5. 0. 5

[H+] = Ka x mmol asam/garamnya

= 2×10^-5 x 2.5/5

= 2×10^-5 x 0.5

= 1×10^-5

pH = – log[H+]

= – log[1×10^-5]

= 5