Calculate the ph of the resulting solution

"Calculate the ph of the resulting solution if 31.0 mL of 0.310 M HCl is added to 21.0 mL of 0.410 M NaOH".

I've tried getting this problem several times and can't come up with the right answer. Please help!

HCl reacts with NaOH to produce a neutral salt NaCl.
The equation:
HCl + NaOH → NaCl + H2O
1mol HCl will react with 1 mol NaOH

Mol HCl in 31.0mL of 0.310M solution = 31/1000*0.31 = 0.00961 mol HCl
Mol NaOH in 21mL of 0.410M solution = 21/1000*0.41 = 0.00861 mol NaOH

When you mix these , the 0.00861 mol NaOH reacts with 0.0861 mol HCl , and 0.0010 mol HCl remains unreacted

You then have 0.0010mol HCl dissolved on 31+21= 52mL of solution

Molarity of HCl solution = 0.001/0.052 = 0.0192M solution

Because HCl is a strong acid , [H+] = [HCl] = 0.0192M
pH = -log [H+]
pH = -log 0.0192
pH = 1.72