Question

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)

Answer 1

Answer:

$pH=-1.37$

Explanation:

We are given that 25 mL of 0.10 M $CH_3COOH$ is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of $CH_3COOH=25mL=0.025 L$

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=$\frac{35}{1000}=0.035 L$

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of $CH_3COOH$=0.10 M

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

Number of moles of H+=$0.10\times 0.025$=0.0025 moles

Number of moles of $OH^-=0.10\times 0.01$=0.001mole

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

Concentration of H+=$\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L$

pH=-log [H+]=-log [4.28$\times 10^{-2}$]=-log4.28+2 log 10=-0.631+2

$pH=-1.37$