Question

4. A solution of 0.200 M NaOH (14.0 mL) is mixed with 42.4 mL of o.170 M HNO,. Assuming that the final volume is the sum of the initial volumes, calculate: (a) the molarity of the Na* cation (b) the molarity of the NO, anion (c) the pH of the final solution (d) the pOH of the final solution

Answer 1

4. when NaOH and HNO3 are mixed

(a) molarity of Na+ cation = 0.2 M x 14 ml/(14 + 42.4) ml = 0.05 M

(b) molarity of NO3- anion = 0.170 M x 42.4 ml/(14 + 42.4) ml = 0.13 M

(c) moles NaOH = 0.2 M x 14 ml = 2.8 mmol

moles HNO3 = 0.170 M x 42.4 ml = 7.208 mmol

after neutralization,

HNO3 remaining = 7.208 - 2.8 = 4.408 mmol

molarity of [H+] in solution = 4.408 mmol/(14 + 42.4) ml = 0.08 M

pH = -log[H+] = -log(0.08) = 1.11

(d) the pOH of the final solution = 14 - pH

                                                   = 14 - 1.11

                                                   = 12.89