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What is pH of mixture prepared by mixing 20 mL 0.07 M NaOH and 13 mL 0.09 M HCl?
Although you are asked about pH, this is a simple limiting reagent question.
When you mix strong acid and strong base they react till one of the reagents runs out. What is left is responsible for pH.
Reaction taking place in the solution is
NaOH + HCl -> NaCl + H2O
Initially there were 20×0.07=1.4 mmol of NaOH and 13×0.09=1.17 mmol of HCl. They react 1:1, so after reaction 1.4-1.17=0.23 mmol of NaOH is left. Final volume of the solution is 20+13=33 mL, thus final concentration of NaOH left is 0.23/33=0.0070 M. pOH=-log(0.0070)=2.15 and pH=14-2.15=11.85.
Note that if there is exactly the same amount of both reactants reaction proceeds to the end - and you are left with perfectly neutral solution of pH=7.00.
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