The answer is pH = 11.85.
The balanced chemical equation is
NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)
SInce NaOH is a strong base and HCl is a strong acid, they will dissociate completely in aqueous solution; likewise, NaCl, being a salt, will dissociate completely in Na+ and Cl− ions, thus giving us the complete ionic equation
Na+(aq)+OH−(aq)+Cl−(aq)+H+3O(aq)→2H2O(aq)+Na+(aq)+Cl−(aq)
SInce Na+ and Cl− ions are present both on the reactants and on the products' side, they are considered to be spectator ions - which leads us to the net ionic equation
H+3O(aq)+OH−(aq)→2H2O(l) - this represents the neutralization reacton between NaOH and HCl.
You could consider this to be a titration problem - let's assume we are titrating 120.0mL of HCL, 0.0150M, with 70.0mL of NaOH, 0.0450M.
We know the volume of HCl to be 120.0mL, which translates to a number of moles of
nHCl=C⋅V=0.0150M⋅120.0⋅10−3L=0.0018 moles
Now we add the 70.0mL of NaOH. The number of NaOH moles added is
nNaOH=C⋅V=0.0450M⋅70.0⋅10−3L=0.00315 moles
Comparing the number of HCl and NaOH moles, which equal the number of H+3O and OH− moles, we can see that we have an excess of OH− moles of
nexcessOH−=nOH−−nH+3O=0.00315−0.0018=0.00135 moles left in the solution. This gives us the concentration of OH−
[OH−]=nexcessVTOTAL=0.00135moles(120.0+70.0)⋅10−3L=0.0071M
(VTOTAL represents the combined volume of the new solution)
So, pOH=−log([OH−])=−log(0.0071)=2.15
Therefore, pH =14−pOH=14−2.15=11.85
