Suppose that 25.0 ml of 0.10 m ch3cooh (aq) is titrated with 0.10 m naoh (aq). what is the ph after the addition of 10.0 ml of 0.10 m naoh (aq)? (notice that the total volume of the solution changes with the addition of 10.0 ml of 0.10 m naoh)

Explanation:

We are given that 25 mL of 0.10 M is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of =0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+==0.0025 moles

Number of moles of =0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=

pH=-log [H+]=-log [4.28]=-log4.28+2 log 10=-0.631+2

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