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

`pH=-1.37`

Step-by-step 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=
`(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* 0.025`=0.0025 moles

Number of moles of
`OH^-=0.10* 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+=
`(0.0015)/(0.035)=4.28* 10^(-2) m/L`

pH=-log [H+]=-log [4.28
`* 10^(-2)`]=-log4.28+2 log 10=-0.631+2

`pH=-1.37`