The ph of a solution prepared by mixing 50.0 ml of 0.125 m naoh and 40.0 ml of 0.125 m hno3 is

Both the acid and the base are known as strong substances which means they totally dissociate into ions in aqueous solution. Mixing the two a salt would be form which is NaNO3. The unreacted substances will be the one contributing to the pH of the solution. 

0.125M HNO3 x 0.04L = 0.005 mol HNO3
0.125M NaOH x 0.05L = 0.00625 mol NaOH

As we can see, all of the HNO3 would react with NaOH so the remaining OH from the unreacted NaOH would be responsible for the pH.

0.00625 - 0.005 / ( 0.0500L + 0.0400L ) = 0.0139 M 

pOH = -log( OH- ) 
pOH = 1.86 
pH = 14.00 - pOH 
pH = 12.14

1. o2(g)as a product

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