Question

(THIS IS A MIXTURE OF ALL OF THE FOLLOWING ACIDS AND BASES MIXED TOGETHER WHERE A COMBINED pH IS WHAT I AM LOOKING FOR)

1. I have 0.500 Liters of water. To that 0.500 Liters, I add the following: a. 0.100 moles HCl b. 0.100 moles HOAc c. 0.100 moles NH4Cl d. 0.100 moles HF e. 0.050 moles NaOH f. 0.050 moles NaOAc g. 0.050 moles Mg(OH)2 What is the pH of the resulting mixture of everything?

Dissociation constants of acids: HOAc Ka=1.78x10-5 H2SO4 Ka1=1x107 Ka2=1.0x10-2 HCNO Ka=2x10-4 H2CO3 Ka1=4.3x10-7 Ka2=5.6x10-11 H2O Ka=1.0x10-14 HIO Ka=2.3x10-11 HNO2 Ka=4.5x10-4 HIO3 Ka = 1.7x10-4 HCl Ka=1x108 HNO3 Ka=1x106 HF Ka = 3.5x10-4 H2C2O4 Ka1 = 6.0x10-2 Ka2 = 6.1x10-5

Dissociation constants of bases: NH3 Kb=1.76x10-5 CH3NH2 Kb=4.4x10-4 NH2OH Kb=1.1x10-8 (CH3)3N Kb=6.5x10-5 C5H5N Kb=1.8x10-9 Mg(OH)2 Kb = 1 x106 NaOH Kb = 1x108

Answer 1

Molarity (M) = (number of moles/ solvent in liter)

a. 0.100 moles of HCl in 0.5 liter water, hence concentration of HCl will be 0.2 M

as HCl is strong acid, so dissociation will be 100% and concentration of HCl will be the concentration of H₃O⁺. HCl + H₂O = H₃O⁺ + Cl^-

Therefore [H₃O⁺]_HCl = 0.2 M

pH can be calculated for weak acid by the following general way...

                                        HA = H⁺ + A^-

Initial concentration     C      0     0

At equilibrium           C-x       x      x       

  K_a = [H⁺][A^- ]/[HA]

            ⁼ x²/ (C-x) = x²/ C     as -x will be very small with respect to C

Hence pH = -log[H⁺] = -log[x] = -log[(Ka * C)^1/2] = -1/2logK_a – 1/2logC ........(1)

Similarly for weak base pOH = -1/2logK_b – 1/2logC..........(2)

b. 0.100 mole HOAc = 0.2 M HOAc

HOAc is the weak acid, so pH will be using equation (1) 5.08

Hence [H₃O⁺]_HOAc = 10^-5.08 M [pH = -log₁₀[H₃O⁺]

c. 0.100 mol NH₄Cl = 0.2 M NH₄Cl, it is a salt of weak base and strong acid , the well known formula for calculating pH = 7-1/2pK_b -1/2logC

therefore, pH = 4.97 and [H₃O⁺]_NH4Cl = 10^-4.97 M

d. 0.100 mole HF = 0.2 M HF. HF is a weak acid, so pH can be calculated using equation 1.

pH =2.08 and [H₃O⁺]_HF= 10^-2.08 M

e. 0.05 molar NaOH =0.1 M NaOH, it is a strong base and it will 100% ionize.

Hence pOH = -log[OH^-] = 1

As it is well known that pH + pOH = 14, so pH = 13

[H₃O⁺]_NaOH= 10^-13 M

f. 0.05 molar NaOAc =0.1 M NaOAc, it is a salt of strong base and weak acid , the well known formula for calculating pH = 7+1/2pK_a + 1/2logC

therefore, pH = 11.24 and [H₃O⁺]_NaOAc = 10^-11.24 M

g. 0.05 molar Mg(OH)₂ =0.1M Mg(OH)₂ it is a strong base and it will 100% ionize.

Mg(OH)₂ = Mg²⁺ + 2OH^-

Hence pOH = -log[2*OH^-] = 0.69 and pH =13.31

Hence [H₃O⁺] _Mg(OH)2 = 10^-13.31 M

So,

[H₃O⁺] _total = [H₃O⁺]_HCl + [H₃O⁺] _HOAC + [H₃O⁺] _NH4Cl +[H₃O⁺] _HF + [H₃O⁺] _NaOH +[H₃O⁺] _NaOAC + [H₃O⁺] _Mg(OH)2

                                 = 0.2 + 10^-5.08 + 10^-4.97 + 10^-2.08 + 10^-13.0 +10^-11.24 +10^-13.31

= 0.2 (approx)

Hence combined pH = -log[0.2] = 0.69