Calculate the ph values of the following solutions: (hint: see special topic i in the study guide and solutions manual.) a 1.0 m solution of acetic acid (pka=4.76) a 0.1 m solution of protonated methylamine (pka=10.7) a solution containing 0.3 m hcooh and 0.1 m hcoo− (pka of hcooh=3.76)

See Explanation

Explanation:

a.pH of 1M HOAc(aq)

                HOAc      ⇄      H⁺   +      OAcˉ

C(eq)         1.0M                 x               x

Ka = [H⁺][OAc⁻]/[HOAc] = x²/1.0M = 1.85x10⁻⁵

=> x = [H⁺] = SqrRt([HOAc]Ka) = SqrRt[(1M)(1.85x10ˉ⁵)] = 4.30x10ˉ³M

=> pH = -log[H⁺] = -log(4.30x10ˉ³) = 2.37

b.pH of 0.10M CH₃NH₃OH(aq)

                CH₃NH₃OH => CH₃NH₃⁺ + OHˉ; Kb = 4.4x10ˉ⁴

C(eq)             0.10M                x              x

=> Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₃] = x²/0.10M

=> x = [OHˉ] = SqrRt([CH₃NH₃OH]Kb) = SqrRt[(0.10M)(4.4x10ˉ⁴)] = 6.63x10ˉ³M

=> pOH = -log[OHˉ] = -log(6.63x10⁻³) = 2.18

=> pH = 14 – pOH = 14 – 2.18 = 11.82

c.pH of 0.30M HOAc/0.10M OAcˉ(aq)

                HOAc      ⇄      H⁺   +      OAcˉ

C(eq)        0.30M               x           0.10M

=> Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]

          = 1.85X10ˉ⁵(0.30M)/(0.10M) = 5.55X10ˉ⁵M

=> pH = -log[H⁺] = -log(5.55x10ˉ⁵) = 4.26

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