Question

Name: Aqueous Ionic Equilibria Test Curve Assignment Directions: Show all work for the following questions and attach any extra sheets used before submitting. This assignment is due on Wednesday, April 24. 1. A buffer is composed of 0.130 M HNO2 and 0.210 M KNO2. The Ka of nitrous acid is 4.6 x 10*. Calculate the pH of the buffer: a. Before the addition of any contaminant. b. After the addition of 50.0 mL of 0.10 M HCl. c. After the addition of 2.07 g of solid NaOH.

Please help

Answer 1

Solution :-

Part a) Calculation of pH before adding any contaminant.

Molarity of acid HNO2 = 0.130 M

Molarity of conjugate base KNO2 = 0.210 M

Ka= 4.6*10^-4

Pka= -log ka

Pka= -log 4.6*10^-4

Pka= 3.34

Henderson equation

pH =pka + log [base]/[acid]

pH= 3.34 + log [0.210]/[0.130]

pH= 3.55

therefore the pH is 3.55

part b) Calculating the pH after addition of 50 ml of 0.10 M HCl

assume initial volume of buffer solution is 1.0 L

initial moles of HNO2 = molarity x volume

                                        = 0.130 mol per L * 1.0 L

                                        = 0.130 mol

Initial moles of KNO2 = 0.210 mol per L * 1.0 L = 0.210 mol

Moles of HCl = 0.10 mol per L * 0.050 L = 0.005 mol

After the reaction moles of acid and base present are

Moles of acid HNO2 = 0.130 mol + 0.005 mol = 0.135 mol

Moles of base KNO2 = 0.210 mol – 0.005 mol = 0.205 mol

pH =pka + log [base]/[acid]

pH= 3.34 + log [0.205]/[0.135]

pH= 3.52

therefore the pH is 3.52

part c) calculating the pH after addition of 2.07 g solid NaOH

initial moles of HNO2 = molarity x volume

                                        = 0.130 mol per L * 1.0 L

                                        = 0.130 mol

Initial moles of KNO2 = 0.210 mol per L * 1.0 L = 0.210 mol

Moles of NaOH = mass/ molar mass

                            = 2.07 g/ 40.0 g per mol

                            = 0.052 mol

After the reaction, moles of acid and base are

Moles of acid HNO2 = 0.130 mol – 0.052 mol = 0.078 mol

Moles of base NO2^- = 0.210 mol + 0.052 mol = 0.262 mol

pH =pka + log [base]/[acid]

pH= 3.34 + log [0.262]/[0.078]

pH= 3.87

therefore the pH is 3.87