Question

Calculate the pH of a solution resulting from the addition of 15.0 mL of a 0.340 M NaOH (aq) to 25.0 mL of 0.250 M HCl (aq)?

Answer 1

Well, we know that `pH=-log_10[H_3O^+]`

We finally get `pH~=1.5`

Explanation:

But first we have to determine the molar equivalence of sodium hydroxide and hydrogen chloride...and this reacts 1:1 according to the following reaction.

`NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)`

`"Moles of NaOH"=15.0*mLxx10^-3*L*mL^-1xx0.340*mol*L^-1=0.00510*mol.`

`"Moles of HCl"=25.0*mLxx10^-3*L*mL^-1xx0.250*mol*L^-1=0.00625*mol.`

And thus there ARE `(0.00625-0.00510)*mol` `HCl` EXCESS in a solution with a volume of `(15.0+25.0)*mL` (the volumes are reasonably assumed to be additive).

`[HCl]=(0.00115*mol)/(40.0xx10^-3*L)=0.02875*mol*L^-1`.

Now (finally)...`pH=-log_10[H_3O^+]=-log_10(0.02875)=-(-1.54)=+1.54`...