Asolution has an initial concentration of 0.0100 m hclo (ka = 3.5×10−8 ) and 0.0300 m naclo. what is the ph after the addition of 0.0030 mol of solid naoh to 1.00 l of this solution? assume no volume change.

When [HClO]= 0.01 M  & [ClO]^- = 0.03M & NaOH = 0.003 mol and
 Ka = 3.5 x 10^-8 
and the equation is:
            HClO + NaOH → Na^+  + ClO^-
initial     0.01     0.003        0.03     0.03
            -0.003 - 0.003      +0.003  + 0.003
Final    0.007        0             0.033      0.033

We can get the PH from this formula:
PH = Pka + ㏒[conjugent base/weak acid]
PH = -㏒Ka + ㏒[Na]/[HClO]
      = - ㏒ 3.5x10^-8 + ㏒(0.033/0.007)
     = 8.13 

D. is the answer hope that you