Analytical Chem

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 2.390. The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10.

I know the following is what needs to be done, but im unsure of how to go about it.

1. Write the pertinent reactions. 2. Write the charge balance equation. 3. Write the mass balance equations. 4. Write the equilibrium constant expressions for each reaction. 5. Count the equations and unknowns. 6. Solve for all unknowns.

  1. 👍
  2. 👎
  3. 👁
  1. Yes but there is an easier way to do it.
    Let S = solubility
    Zn(CN)2==> Zn^2+ + 2CN^-,then + H^+ =>HCN
    ..solid....S.......2S
    Ksp = (Zn^2+)(CN^-)^2

    For HCN ==> H^+ + CN^- You should confirm all of this.
    Ka = 6.2E-10 = (H^+)(CN^-)/(HCN)
    If pH 2.39 then (H^+) = about 4E-3 but you can do it more accurately than that.
    Then (4E-3)(CN^-)/(HCN) = 6.2E-10=(4E-3)(CN^-)/(HCN) and solve for (HCN) = 6.56E6(CN^-) and substitute this into the below equation.

    You know (Zn^2+) = S and you know
    2S = (CN^-) + (HCN)
    2S = (CN^-) + 6.56E6 = essentially 6.56E6.
    S = 3.28E6(CN^-) and
    (CN^-) = S/3.28E6
    Now substitute S for Zn and S/3.28E6 for CN and solve for S.

    1. 👍
    2. 👎
  2. I am a bit confused at the substitution part at the end. Can you please break it down further so I can find the concentrations?

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. Chemistry

    Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (Ka (HCN) = 4.9 × 10–10) My answer is 9.52, but im not sure.

  2. Chemistry

    1.33dm³ of water at 70°c are saturated by 2.25moles of Pb(NO3)2, and 1.33dm³ of water at 18°c are saturated by 0.53mole of the same salt. If 4.50dm³ of the saturated solution are cooled from 70°c to 18°c.calculate the

  3. chemistry

    calculate the pH of a 0.50 m solution of HCN. Ka for HCN is 4.9*10^-10 (The chemical equation for this.)

  4. chemistry

    How do you calculate the theoretical cell voltage? 1) Zn|Zn^+2 (1M)| Cu|Cu^+2 (1M) okay so by looking at example this example Calculate the emf (voltage) for the following reaction: Zn(s) + Fe2+ → Zn2+ + Fe(s) Write the 2 half

  1. Chemistry

    A student places a zinc electrode in a 0.80 M Zn2+(aq) solution which is connected by an electrolyte to a 1.30 M Ag+ (aq) solution containing a silver electrode. (Note that the solution concentrations are not standard). Determine

  2. chem

    Calculate the pH of 0.20 M NaCN solution. NaCN ---> Na+ + CN- CN- + H20+ ---> HCN+ + OH- Initial conc. of CN- = 0.20 mol/L change = -x equillibrium = 0.20-x HCN equill. = +x OH equill. = 1x10^-7+x Ka= 6.2 x 10^-10 Kb = KW/Ka =

  3. Quantitative Chemistry

    Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 2.92. Ksp (Zn(CN)2) = 3.0 × 10-16; Ka (HCN) = 6.2 × 10-10. Completely lost!

  4. Chemistry

    Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 × 10-59, and dissociates according to Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq) Solid hydroxylapatite is dissolved in water to form a

  1. Chemistry

    Consider the voltaic cell: Zn(s) / Zn2+(aq) // H+(aq) / H2(g) / Pt(s) Calculate the value of Ecell, in V, when [Zn2+] =1.00 M and [H+] = 0.00100 M DATA: Zn2+(aq) +2 e¯ ---> Zn(s);Eo = - 0.76 V 2 H+(aq) + 2 e¯ ---> H2(g);Eo =

  2. chemistry

    Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (Ka(HCN) = 4.9 × 10^–10)

  3. chemistry

    What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10-10? Rate my answer please For your understanding: HCN ⇌ H+ + CN-, Ka = 6.2x10^-10 = [H+]*[CN-]/[HCN] Notice the H+ and CN- ions are generated in pairs

  4. Chemistry

    Calculate the pH of the solution resulting from the addition of 75.0 mL of 0.15 M KOH to 35.0 mL of 0.20 M HCN (Ka (HCN) = 4.9 × 10–10).

You can view more similar questions or ask a new question.