What is the $pH$ of an aqueous solution prepared from a $600mg$ mass of $NaOH$ dissolved in $75mL$ of water?

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Answer 1 · 2016-10-27T08:38:39

$pH~=13$

We need to find $[HO^-]$ $=$ $"Moles of hydroxide"/"Volume of water"$ $=$ $(0.600*g)/(40.00*g*mol^-1)xx1/(75.0xx10^-3L)$ $=$ $0.20*mol*L^-1$ .

Now $pOH$ $=$ $-log_10[HO^-]$ $=$ $-log_10(0.20)$ $=$ $-(-0.70)$ $=$ $0.70$ .

But we know (or should know) that in water at $298*K$ , $pH+pOH=14$ .

Thus $pH=14-0.70=?????????$

What is the pHpH of an aqueous solution prepared from a 600*mg600*mg mass of NaOHNaOH dissolved in 75*mL75*mL of water?