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## CHM 112 Section 1

Homework Assignment #19
##### Due March 27, 2006 by 8:05 a.m.

Practice (not to be turned in): Problems 15.81, 15.133

A solution is prepared at 25 °C using sodium phosphate and sodium hydrogen phosphate. The concentration of the sodium phosphate equals the first two digits of your Student ID Number divided by 100 in units of molarity. The concentration of the sodium hydrogen phosphate equals the eighth and ninth digits of your Student ID Number divided by 100 in units of molarity.

a) Find the pH of the buffer solution.

b) 1.0 g of HCl(g) is added to 1.00 L of the buffer solution. Find the new pH.

c) 0.50 g of NaOH(s) is added to 1.00 L of the buffer solution. Find the new pH.

a) The unperturbed buffer solution:

 Na3PO4(aq) 3 Na+(aq) + PO43–(aq) Na2HPO4(aq) 2 Na+(aq) + HPO42–(aq) HPO42–(aq) + H2O(l) H3O+(aq) + PO43–(aq) Initial [HPO42– ]init 0 [PO43– ]init Change –x + x + x Equilibrium [HPO42– ]init – x x [PO43– ]init + x

[HPO42– ]init = (eighth and ninth digits of Student ID Number)/100 M

[PO43– ]init = 0.10 M

Approximation is allowed in all cases, so

4.2×10–13 = (x)(0.10) / [HPO42– ]init

x = [H3O+]e

pH = –log[H3O+]e to 2 digits past the decimal place.

b) 1.0 g HCl added to 1.00 L of the buffer solution:

The molar mass of HCl = (35.5 + 1.0) g/mol = 36.5 g/mol

The added concentration of HCl = 1.0/36.5 = 0.027 M

 Na3PO4(aq) 3 Na+(aq) + PO43–(aq) Na2HPO4(aq) 2 Na+(aq) + HPO42–(aq) HPO42–(aq) + H2O(l) H3O+(aq) + PO43–(aq) HPO42–(aq) HCl(g) + PO43–(aq) Initial [HPO42– ]init + 0.027 0 0.10 – 0.027 Change – x + x + x Equilibrium [HPO42– ]init + 0.027 – x x 0.10 – 0.027 + x

[HPO42– ]init = (eighth and ninth digits of Student ID Number)/100 M

[PO43– ]init = 0.10 M

Approximation is allowed in all cases, so

4.2×10–13 = (x)(0.10 – 0.027) / ([HPO42– ]init + 0.027)

x = [H3O+]e

pH = –log[H3O+]e to 2 digits past the decimal place.

c) 0.50 g NaOH added to 1.00 L of the buffer solution:

The molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol

The added concentration of NaOH = 0.50/40.0 = 0.013 M

 Na3PO4(aq) 3 Na+(aq) + PO43–(aq) Na2HPO4(aq) 2 Na+(aq) + HPO42–(aq) HPO42–(aq) + H2O(l) H3O+(aq) + PO43–(aq) HPO42–(aq) + NaOH(aq) H2O(l) + Na+(aq) + PO43–(aq) Initial [HPO42– ]init – 0.013 0 [PO43–]init + 0.013 Change – x + x + x Equilibrium [HPO42– ]init – 0.013 – x x [PO43–]init + 0.013 + x

[HPO42– ]init = (eighth and ninth digits of Student ID Number)/100 M

[PO43– ]init = 0.10 M

Approximation is allowed in all cases, so

4.2×10–13 = (x)(0.10 + 0.013) / ([HPO42– ]init – 0.013)

x = [H3O+]e

pH = –log[H3O+]e to 2 digits past the decimal place.