Question

pH of a solution made by mixing 50 mL of 0.2 M NH4Cl and 75 mL of 0.1 M NaOH is

Answer 1

Here,pKb of the base was required. However, the pKb for NaOH is 0.2(very strong base).
Now, here two basic solutions are mixed one is basic salt and another is base so a basic buffer is produced.
So, we have, 
pOH=pKb+log[salt][base]

Now, final solution volume of NaOH=50+75 =  125 ml 
So, M1V1= M2V2
0.1 x50 = M2 x 125
M2 =0.04M
Similarly, final volume of ​NH​4Cl=25 ml
So, M1V1= M2V2
0.2 x50 = M2 x 25
M2 =0.4M
Hence, pOH =0.2 +log[0.4]/[0.04]= 0.2+1=1.2
Hence, pH = 14-pOH = 14-1.2= 12.8