Question

In: Chemistry

In the titration of 25.0 mL of 0.1 M CH3COOH with 0.1 M NaOH, how is...

In the titration of 25.0 mL of 0.1 M CH3COOH with 0.1 M NaOH, how is the pH calculated after 8 mL of titrant is added?

a	The pH is 14.

b	The pH is calculated using the H-H equation for a buffer solution, using the ratio of the concentrations of the weak base and the weak acid, and the pKaof the acid.

c	The pH is 1.

d	The pH is calculated by determining the concentration of weak conjugate acid present in the solution, using an ICE table to calculate the proton concentration present after hydrolysis, and taking the negative log of the result.

e	The pH is based on the concentration of protons present in the solution, which is equal to the original concentration of the base.

f	The pH = pKa of the acid.

g	The pH is 7.

h	The pH is calculated by determining the concentration of weak conjugate base present in the solution, using an ICE table to calculate the hydroxide ion concentration present after hydrolysis, subtracting pOH from 14, and taking the negative log of the result.

i	The pH is calculated by determining the concentration of leftover hydroxide ions in the solution, subtracting pOH from 14, and taking the negative log of the result.

Expert Solution

CH3COOH = 25.0mL of 0.1M

number of moles of CH3COOh = 0.1Mx0.025L= 0.0025 moles

NaOH= 8 mL of 0.1M

number of moles of NaOH= 0.1Mx0.008L= 0.0008 moles

Pka of CH3COOH = 4.76

CH3COOH + NaOH ----------------- CH3COONa + H2O

0.0025 0.0008 0 0

- 0.0008 -0.0008 + 0.0008 +0.0008

0.0017 0 + 0.0008

number of moles of CH3COOH = 0.0017 moles

number of moles of CH3COONa = 0.0008 moles

PH= PKa + log[slat/acid]

PH= 4.76 + log(0.0008/0.0017)

PH= 4.43

The answer is b.

queen_honey_blossom answered 2 years ago

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