Question

A 1.00 liter solution contains 0.42 M acetic acid and 0.55 M sodium acetate. If 0.280 moles of sodium hydroxide are added to this system, indicate whether the following statements are true or false. (Assume that the volume does not change upon the addition of sodium hydroxide.) A. The number of moles of CH,COOH will decrease. B. The number of moles of CH3COO will increase. C. The equilibrium concentration of H20* will increase. D. The pH will increase. E. The ratio of (CH,COOH]/(CH,CO0] will increase. Submit Answer Retry Entire Group 4 more group attempts remaining

Answer 1

Given:-

volume of solution (V) = 1.00 L

molarity of CH₃COOH i.e [CH₃COOH] = 0.42 M

molarity of CH₃COONa i.e [CH₃COONa] = 0.55 M

No. of moles of NaOH (n_NaOH) = 0.280 mol

As we know that

pKa of CH₃COOH = 4.76

Also we know that

molarity of compound (M) = No. of moles of compound (n) / volume of solution (V) in liter

No. of moles of compound (n) = molarity of compound (M) $\times$ volume of solution (V) in liter

therefore

No. of moles of CH₃COOH (n_CH3COOH) =  molarity of CH₃COOH (M) $\times$ volume of solution (V) in liter

No. of moles of CH₃COOH (n_CH3COOH) = 0.42 M $\times$ 1.00 L

No. of moles of CH₃COOH (n_CH3COOH) = 0.42 mol

similarly

No. of moles of CH₃COONa (n_CH3COONa) = molarity of CH₃COONa (M) $\times$ volume of solution (V) in liter

No. of moles of CH₃COONa (n_CH3COONa) = 0.55 M $\times$ 1.00 L

No. of moles of CH₃COONa (n_CH3COONa) = 0.55 mol

So according to the Henderson hasselbalch equation

pH = pKa + log [salt] / [acid]

pH = pKa + log [n_CH3COONa] / [n_CH3COOH]

pH =4.76 + log (0.55 / 0.42)

pH =4.76 + log (1.31)

pH = 4.76 + 0.1172

pH = 4.88 ( before addition of NaOH)

As we know that when 0.280 moles of NaOH is added to the buffer solution the following reaction would occur

CH₃COOH + NaOH   $\rightarrow$ CH₃COONa + H₂O

Initial  0.42 mol 0 0.55 0

Change - 0.280 mol - 0.280 mol + 0.280 mol   0.280 mol

therefore

Decreased no. of moles of CH₃COOH (n'_CH3COOH) =  0.42 mol  - 0.280 mol

Decreased no. of moles of CH₃COOH (n'_CH3COOH) = 0.14 mol

and

Increased no. of moles of CH₃COONa (n'_CH3COONa) =  0.55 mol  + 0.280 mol

Increased no. of moles of CH₃COONa (n'_CH3COONa) = 0.83 mol

So according to the Henderson hasselbalch equation

pH = pKa + log [salt] / [acid]

pH = pKa + log [n_CH3COONa] / [n_CH3COOH]

pH =4.76 + log (0.83 / 0.14)

pH =4.76 + log (5.93)

pH = 4.76 + 0.7731

pH = 5.53 ( After addition of NaOH)

Therefore as mentioned above the addition of NaOH brings about following change which are given below:

1- The number of moles of CH₃COOH will decrease.

2- The number of moles of CH₃COONa will increase.

3- The equilibrium concentration of H₃O+ will decrease.

4- The pH will increase.

5- The ratio of [CH₃COOH] / [CH₃COONa] will decrease as follows:-

Before addition of NaOH

ratio of [CH₃COOH] / [CH₃COONa] = 0.42 / 0.55 = 0.764

After addition of NaOH

ratio of [CH₃COOH] / [CH₃COONa] = 0.14 / 0.83 = 0.169