Question #31b0a

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Answer 1 · 2017-03-22T05:59:18

Well, this is a basic solution; because it contains MORE of the acidium ion than the hydroxide ion.........

We interrogate the equilibrium:

$2H_2O rightleftharpoons H_3O^+ + HO^-$

For which $K_w=10^(-14)=[H_3O^+][HO^-]$

Taking $log_10$ of both sides:

$log_(10)10^-14=log_10[H_3O^+]+log_10[HO^-]$

And on rearrangement, $14=-log_10[H_3O^+]-log_10[HO^-]$ ,

i.e. $14=pH+pOH$ .

$"Concentration of sodium hydroxide"$

$=("Moles of sodium hydroxide")/("Volume of solution")$

$=$ $((0.800*g)/(39.998*g*mol^-1))/(2.00*L)=1.00xx10^-2*mol*L^-1$

Now, clearly, we can take the $pH$ of this solution, given that we know that $pH+pOH=14$ , and $pOH=-log_10[HO^-]$
$=-log_10(1.00xx10^-2)=2.0$

Thus $pH=14-2=12$ ......clearly an alkaline solution,