# Solutions

- Page ID
- 11056

**1. Calculate the equilibrium concentrations and pH for a 0.20 M proprionic acid solution.**

**Complete the equilibrium and express the equilibrium concentrations in terms of a single unknown 'x'.**

HC _{3}H_{5}O_{2 (aq)} | H ^{+}_{(aq)} | + C _{3}H_{5}O_{2}^{- }_{(aq)} | ||

initial concentration | 0.2 M | 0 | 0 | |

change | - x | + x | + x | |

equilibrium concentration | 0.2 - x | x | x |

**Write the equilibrium expression (K _{a}) and express the K_{a} in terms of the equilibrium concentrations.**

[H^{+}][C_{3}H_{5}O_{2}^{-}] | (x)(x) | |||

K _{a} = | ^{_______________} | = | ^{___________} | = 1.3 x 10 ^{-5} |

[HC_{3}H_{5}O_{2}] | (0.2 - x) | (from table) |

**Solve for x; since the initial concentration is greater than 0.1 and the K _{a} less than 10^{-4}, we can drop the '- x' term.**

x ^{2} | solving for 'x' gives | ||

^{_______} | = 1.3 x 10 ^{-5} | x = 1.61 x 10 ^{-3} | |

0.2 | |

**Solve for the equilibrium concentrations by substituting the value of 'x' into the equilibrium concentrations: [H ^{+}] = [C_{3}H_{5}O_{2}^{-}] = x, [HC_{3}H_{5}O_{2}] = 0.2 - x**

**[H ^{+}] = [C_{3}H_{5}O_{2}^{-}] = 1.61 x 10^{-3} M, [HC_{3}H_{5}O_{2}] = 1.98 x 10^{-1} M;**

**pH = - log [H ^{+}] = - log(1.61 x 10^{-3})**

**pH = 2.79**

**2. Calculate the equilibrium concentrations and pH for a 0.20 M carbonic acid solution.**

** Compare the answer to problem 1. Note which has the higher K _{a} and lower pH.**

**This is done the same way as problem 1, using a K**

_{a}value of 4.2 x 10^{-7} H _{2}CO_{3 (aq)} | H ^{+}_{(aq)} | + HCO _{3}^{- }_{(aq)} | ||

initial concentration | 0.2 M | 0 | 0 | |

change | - x | + x | + x | |

equilibrium concentration | 0.2 - x | x | x |

[H^{+}][HCO_{3}^{-}] | (x)(x) | |||

K _{a} = | ^{_______________} | = | ^{___________} | = 4.2 x 10 ^{-7} |

[H_{2}CO_{3}] | (0.2 - x) | (from table) |

x ^{2} | solving for 'x' gives | ||

^{_______} | = 4.2 x 10 ^{-7} | x = 2.90 x 10 ^{-4} | |

0.2 | |

**[H ^{+}] = [HCO_{3}^{-}] = 2.90 x 10^{-4} M, [H_{2}CO_{3}] = 2.0 x 10^{-1} M; pH = 3.54**

**The higher the K _{a}, the lower the pH for the same concentration of weak acid. The higher K_{a} value tells us we have more dissociation of the weak acid, giving a greater [H^{+}].**

**3. Calculate the K _{a} for a 0.3 M solution of HA (weak acid) if the pH = 3.65**

**First calculate the [H**

^{+}] concentration from the pH**[H**

^{+}] = 10^{-pH}; [H^{+}] = 10^{-3.65}; [H^{+}] = 2.24 x 10^{-4}**since 1:1 mole ratio**

**[H ^{+}] = [A^{-}]; [HA] = 0.3 - [H^{+}]**

**Substitute into the K _{a }expression**

[H^{+}][A^{-}] | (2.24 x 10 ^{-4})(2.24 x 10^{-4}) | ||

K _{a} = | ^{________} | = | ^{________________________} |

[HA] | (0.3 - 2.24 x 10 ^{-4}) |

**K _{a}= 1.67 x 10^{-7}**

**4. Calculate the pH for a 0.2 M pyridine solution.**

C _{5}H_{5}N_{ (aq)} | OH ^{-}_{(aq)} | + C _{5}H_{5}NH^{+ }_{(aq)} | ||

initial concentration | 0.2 M | 0 | 0 | |

change | - x | + x | + x | |

equilibrium concentration | 0.2 - x | x | x |

[OH^{-}][C_{5}H_{5}NH^{+}] | (x)(x) | |||

K _{b} = | ^{________________} | = | ^{___________} | = 2.0 x 10 ^{-9} |

| (0.2 - x) | (from table) |

**Solve for 'x'**

x ^{2} | solving for 'x' gives | ||

^{_______} | = 2.0 x 10 ^{-9} | x = 2.0 x 10 ^{-5} | |

0.2 | |

**[OH ^{-}] = [C_{5}H_{5}NH^{+}] = 2.0 x 10^{-5} M, [C_{5}H_{5}N] = 2.0 x 10^{-1} M**

**Using the following equations: pOH = - log [OH ^{-}]; pH + pOH = 14**

**pOH = 4.70, pH = 9.30**

**5. Calculate the pH of a 0.34 M HCl solution.**

**pH = 0.47, strong acid, 100 % ionization; 1:1 mole ratio, [HCl] = [H ^{+}]**