Question

PS11.6. Determine the pH of a buffer prepared by mixing 0.675 moles of HC2H3O2 and 0.575 moles of NaC2H3O2 in enough water to give 1.00 liter of solution.

Calculate the pH when

a) 0.0500 mol of HCl is added to the buffer.

b) 0.0500 mol of NaOH is added to the original buffer

c) 1 liter of water is added to the original buffer

d) 5.00 mL of a 1.00 M HNO3 solution is added to the original buffer

Answer 1

PS11.6. Determine the pH of a buffer prepared by mixing 0.675 moles of HC2H3O2 and 0.575 moles of NaC2H3O2 in enough water to give 1.00 liter of solution.

Calculate the pH when

a) 0.0500 mol of HCl is added to the buffer.

b) 0.0500 mol of NaOH is added to the original buffer

c) 1 liter of water is added to the original buffer

d) 5.00 mL of a 1.00 M HNO3 solution is added to the original buffer

1)For HC2H3O2(acetic acid) / NaC2H3O2(sodium acetate) buffer we have

[salt]=[ NaC2H3O2] =0.575 M/L,

[acid] = [HC2H3O2] = 0.675 M/L.

For acetic acid, pKa = 4.76

Henderson’s equation for this acidic buffer is,

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

pH= pKa + log{[ NaC2H3O2]/ [ HC2H3O2]}

pH = 4.76 + log (0.575/0.675)

pH =4.76 + log(0.8519)

pH = 4.76 + (-0.07)

pH = 4.69

pH of given buffer is 4.69.

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a)Change in pH on addition of 0.05 M HCl.

HCl is strong acid and will increase concentration of H⁺ ions in solution i.e. lower concentration of conjugate base (salt) and increase concentration od acid.

New

[salt]=[ NaC2H3O2] = 0.575 M/L – (0.05M x1L) = 0.525 M

[acid] = [HC2H3O2] = 0.675 M/L + (0.05M x1L) = 0.725 M

With these new concentrations again by Henderson’s equation,

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

pH= pKa + log{[ NaC2H3O2]/ [ HC2H3O2]}

pH = 4.76 + log (0.525/0.725)

pH =4.76 + log(0.7241)

pH = 4.76 + (-0.14)

pH = 4.62

pH of given buffer is 4.62.

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B)Change in pH on addition of 0.05 M NaOH.

NaOH is strong base and will increase concentration of OH^- ions in solution i.e. lower concentration of conjugate acid and increase concentration of base.

New

[salt]=[ NaC2H3O2] = 0.575 M/L + (0.05M x1L) = 0.625 M

[acid] = [HC2H3O2] = 0.675 M/L – (0.05M x1L) = 0.625 M

With these new concentrations again by Henderson’s equation,

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

pH= pKa + log{[ NaC2H3O2]/ [ HC2H3O2]}

pH = 4.76 + log (0.625/0.625)

pH =4.76 + log(1)

pH = 4.76 + 0

pH = 4.76

pH of given buffer is 4.76.

pH ie numerically equal to pKa (this we have when acid and salt concentration is same)

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c) Addition of water to buffer do not change its pH.

Because addition of water alters concentration of acid and conjugate base equally so there ration will remains same hence log of ratio. Ultimately pH remains same.

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d) Addition of 5 mL 1.0M HNO3

HNO3 is strong acid and ionizes completely. Addition of HNO3 will increase acid concentration but lowers conjugate base concentration.

New,

[salt]=[ NaC2H3O2] = 0.575 M/L – (1M x 5 x 10^-3L) = 0.57 M

[acid] = [HC2H3O2] = 0.675 M/L + (0.05M x1L) = 0.68 M

With these new concentrations again by Henderson’s equation,

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

pH= pKa + log{[ NaC2H3O2]/ [ HC2H3O2]}

pH = 4.76 + log (0.57/0.68)

pH =4.76 + log(0.8382)

pH = 4.76 + (-0.077)

pH = 4.683

pH of given buffer is 4.683.

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