Hello,
The first question can easily be solved, since NaOH fully dissociates. But I can not say the same for acetic acid, you'll need to give the dissociation constant for it. I remember that it is 1,8*10^{-5} mol^2L^{-2}, and will solve by using this value. If I'm mistaken, please replace it with the correct one.
Firstly, let's look at NaOH:
\underbrace{NaOH} \rightarrow \underbrace{Na^+} + \underbrace{OH^-}
\overbrace{0,01-x} \rightarrow \overbrace{x} + \overbrace{x}
Since dissociation constant for NaOH is extremely big, we never need to calculate the {0,01-x}; it has a very very close value to 0,01. So we can easily assume that the [OH^-] is 0,01, the same goes for [Na^+].
As pH is the negative logarithm of [H^+] or \frac{10^{-14}}{[OH^-]}, we'll use whichever is suitable. Since we know the hydroxide concentration, let's use this:
pH=-\log(\frac{10^{-14}}{0,01})=12.
For acetic acid, we'll consider the x values as it is not dissociated 100% in water.
\underbrace{CH_3COOH} \rightarrow {CH_3COO^-}+{H^+}
\overbrace{(0,035-x)}\rightarrow
We are given that \frac{x^2}{(0,035-x)}=1,8*10^{-5}. Then it's easy to find x, either by omitting it or by solving a two-unknown equation with \Delta=b^2-4ac and x_1=\frac{-b-\sqrt{\Delta}}{2a} and x_2=\frac{-b+\sqrt{\Delta}}{2a}. Note that only one root gives a valid value, just omit the other.
I will not consider x, if you really wonder, you may not omit it and solve the two-unknown equation. When we omit it, we'll find that x=\sqrt{0,035*1,8*10^{-5}}=7,94*10^{-4}, hence we find the pH to be 3,1.
Regards,
chem_tr
