What is the $p H$ $pH$ of an aqueous solution in which $[H O^{-}] = 0.041 \cdot m o l \cdot L^{- 1}$ $[HO^-]=0.041molL^-1$ ?

Question

1907 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-10T16:32:05

$p H = 12.6$ $pH=12.6$

We know that in aqueous solution under standard conditions.....

$p H + p O H = 14$ $pH+pOH=14$ .

And $p O H = - {log}_{10} [H O^{-}]$ $pOH=-log_(10)[HO^-]$ ; so here $p O H = - {log}_{10} (0.041) =$ $pOH=-log_10(0.041)=$

$- {log}_{10} 0.041 = - (- 1.39) = 1.39$ $-log_(10)0.041=-(-1.39)=1.39$ .

And thus $p H = 14 - p O H = 12.6$ $pH=14-pOH=12.6$

So is this solution acid or alkaline?

What is the pHpHpH of an aqueous solution in which [HO^-]=0.041*mol*L^-1[HO−]=0.041⋅mol⋅L−1[HO^-]=0.041*mol*L^-1?