Question

Hitunglah ph larutan campuran 1ml (nh4)2so4 0,2M dengan 40ml nh4oh 0,2M (kb=nh4oh= 2x10 pangkat -5)

Answer 1

Verified answer

[OH-] = Kb (mol nh4oh/mol (nh4)2so4)
= 2×10^-5 ((0,2×40)/(0,2×1))
= 2×10^-5 (8/0,2)
= 8×10^-4

pOH = -log[OH-]
= -log(8×10^-4)
= 4-log8
pH = 14-(4-log8)
= 10+log8

Answer 2

Materi : Asam Basa
Kata kunci : penyangga basa, buffer basa

Gunakan rumus lar. penyangga basa

[OH-] = 2×10^-5 × (40×0.2)/(1×0.2)
= 8×10^-4 M

pH = 14 + log 8×10^-4
= 10 + log 8
= 10.9



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