suatu basa lemah boh 0,01 m mempunyai nilai ph =9 -log 2 hitunglah nilai kb basa tersebut

pH+pOH = 14
9-log 2 + pOH = 14
pOH = 14-(9-log2)
pOH = 5+ log2
OH- = 2×10^5

[tex] {oh}^{ - } = \sqrt{kb \: \times mb} \\ 2 \times {10}^{ - 5} = \sqrt{kb \: \times 1 \times {10}^{ - 2} } \\ 4 \times {10}^{ - 10} = kb \times 1 \times {10}^{ - 2} \\ kb \: = \frac{4 \times {10}^{ - 10} }{1 \times {10}^{ - 2} } \\kb \: = 4 \times {10}^{ - 8} [/tex]