What is the pH of a solution that has a [OH-] = 0.0033 M?

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Answer 1 · 2018-04-19T01:31:49

$p H = 11.52$ $pH=11.52$

Here $[O H^{-}] = 33 \times 10^{- 4} M$ $[OH^-]=33xx 10^(-4)M$

$\Rightarrow p O H = - {log}_{10} [O H^{-}]$ $rArr pOH=-log_10[OH^-]$

$\Rightarrow p O H = - {log}_{10} [33 \times 10^{- 4}]$ $rArr pOH=-log_10[33xx10^(-4)]$

$\Rightarrow p O H = - {log}_{10} (33) - {log}_{10} (10^{- 4})$ $rArr pOH=-log_10(33)-log_10(10^(-4))$

$\Rightarrow p O H = - 1.52 + 4.00$ $rArr pOH=-1.52+4.00$

$\Rightarrow p O H = 2.48$ $rArr pOH=2.48$

We know that $p H + p O H = p K_{w}$ $pH+pOH=pK_w$ ............ ${p K_{w} = 14}$ ${pK_w=14}$

$\Rightarrow p H + 2.48 = 14$ $rArr pH+2.48=14$

$:.pH=11.52$