Hitung ph dari larutanbNH4cl 0.008 jika diketahui kb1*10^-5

[tex]m = 0.008 = 8 \times {10}^{ - 3} \\ kb = 1 \times 10 {}^{? - 5} = {10}^{ - 5} \\ | {h}^{ + } | = \sqrt{kb\times m} \\ = \sqrt{ {10}^{ - 5} \times 8 \times 10 {}^{ - 3} } \\ = \sqrt{8 \times {10}^{ - 8} } \\ = {10}^{ - 4} \sqrt{8} \\ ph = - log| {h}^{ + } | \\ = - log {10}^{ - 4} \sqrt{8} \\ = 4 - log \sqrt{8} [/tex]