Calculate the concentration of hydronium tons in: (a) a solution that is 0.20M H

Question

Calculate the concentration of hydronium tons in: (a) a solution that is 0.20M HBrO(aq) and 0.10M KBrO(aq); (b) a solution that is 0.010M (CH_3)_2NH(aq) and 0.150M (CH_3)_2 NH_2Cl(aq); (c) a solution that is 0.10M HBrO(aq) and 0.20M KBrO(aq); (d) a solution that is 0.020M (CH_3)_2NH(aq) and 0.030M (CH_3)_2NH_2Cl(aq). Sodium hypochlorite, NaClO, is the active ingredient in many bleaches. Calculate the ratio of the concentrations of ClO^- and HClO in n bleach solution having a pH adjusted to 6.50 by using strong acid or strong base. The molarity of CrO_4^2- in a saturated Tl_2CrO_4 solution is 6.3 times 10^-5 mol/L. What is the K_sp of Tl_2CrO_4? Determine the pH required for the onset of precipitation of Ni(OH)_2 from 0.010M NiSO_4(aq). (K_sp = [Ni^2+][OH^-]^2 = 6.5 times 10^-18) Consider the two equilibria: CaF_2(s) Ca^2+(aq) + 2 F^-(aq), K_sp = 4.0 times 10^-11 F^-(aq) + H_2O(l) HF(aq) + OH^-(aq), K_b = 2.9 times 10^-11 (a) Write the chemical equation for the overall equilibrium and determine the corresponding equilibrium constant (b) Determine the solubility of CaF_2 as a function of pH. (c) Determine the solubility of CaF_2 at pH = 7.0.

Explanation / Answer

1) Hydronium ion = H3O+

a) pKa of HOBr = 8.6

From Henderson-Hasselbalch equation,

pH = pKa + log [conjugate base]/[acid]

= 8.6 + log [KBrO]/[HBrO]

= 8.6 + log [0.1 M]/[0.2 M]

= 8.3

pH = 8.3

Then,

[H+] = 10-pH = 10-8.3 = 5.0 x 10-9 M

Therefore,

[H3O+] = [H+] = 5.0 x 10-9 M

b)

pKb of (CH3)2NH = 3.23

pOH = pKb + log [conjugate acid]/[base]

= 3.23 + log [(CH3)2NH2Cl] / [(CH3)2NH]

= 3.23 + log [ 0.15 M] /[0.01 M]

= 4.4

pOH = 4.4

pH = 14 -pOH = 14 - 4.4 = 9.6

Then,

[H+] = 10-pH = 10-9.6 = 2.5 x 10-10 M

Therefore,

[H3O+] = [H+] = 2.5 x 10-10 M

c)

pKa of HOBr = 8.6

From Henderson-Hasselbalch equation,

pH = pKa + log [conjugate base]/[acid]

= 8.6 + log [KBrO]/[HBrO]

= 8.6 + log [0.2 M]/[0.1 M]

= 8.9

pH = 8.9

Then,

[H+] = 10-pH = 10-8.9 = 1.25 x 10-9 M

Therefore,

[H3O+] = [H+] = 1.25 x 10-9 M

d)

pKb of (CH3)2NH =3.23

pOH = pKb + log [conjugate acid]/[base]

= 3.23 + log [(CH3)2NH2Cl] / [(CH3)2NH]

= 3.23 + log [ 0.03 M] /[0.02 M]

= 3.4

pOH = 3.4

pH = 14 -pOH = 14 - 3.4 = 10.6

Then,

[H+] = 10-pH = 10-10.6 = 2.5 x 10-11 M

Therefore,

[H3O+] = [H+] = 2.5 x 10-11 M

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