Question

In: Chemistry

I have a mixture consisting of 0.180 M nitrous acid, and 0.150 M sodium nitrite. NOTE:...

I have a mixture consisting of 0.180 M nitrous acid, and 0.150 M sodium nitrite. NOTE: pKa (nitrous acid) = 3.14. 7.

First, I add 2.00 mL of 0.500 M HCl to 20.00 mL of the original (sodium nitrite / nitrous acid) mixture. Assuming these volumes are additive, make an ICE table and:

a) Write the corresponding balanced reaction, specifying ALL of its relevant reactant and product components.

b) Calculate how many moles of each relevant reaction component are initially present.

c) Calculate the change in the moles of each relevant reaction component.

d) Calculate the moles of each relevant reaction component at equilibrium.

e) Calculate the new molar concentration of nitrite anions.

f) Calculate the new molar concentration of nitrous acid.

g) Calculate the new pH of the resulting mixture.

Solutions

Expert Solution

(a): HCl reacts with sodium nitrite to form nitrous acid and aq.NaCl. The balanced reaction is

NaNO2(aq) + HCl(aq) ------- > HNO2(aq) + NaCl(aq)

(b): Initial moles of nitrous acid, HNO2 = MxV = 0.180 mol/L x 0.020 L = 0.0036 mol

Initial moles of sodium nitrite, NaNO2 = MxV = 0.150 mol/L x 0.020 L = 0.003 mol

initial moles of HCl added = MxV = 0.500 mol/L x 0.002 L = 0.001 mol

(c) ---------- NaNO2(aq) + HCl(aq) ------- > HNO2(aq) + NaCl(aq)

initial mol:0.003 mol, ---- 0.001 mol ------- 0.0036 mol

--change: - 0.001 mol, - 0.001 mol, ----- + 0.001) mol

eqm.mol: (0.003 - 0.001), (0.001 - 0.001), (0.0036 + 0.001) mol

Hence moles of NaNO2 decreases by 0.001 mol.

HCl is completely neutralized.

Moles of HNO2 increases by 0.001 mol

(d) equilibrium mol of NaNO2 = (0.003 - 0.001) mol = 0.002 mol

equilibrium mol of HCl = (0.001 - 0.001) mol = 0.00 mol

equilibrium mol of HNO2 = (0.0036 + 0.001) mol = 0.0046 mol

(e): Total volume at equilibrium, Vt = 20.00 + 2.00 = 22.00 mL = 0.0220 L

[NaNO2(aq)] = 0.002 mol / 0.0220 L = 0.091 M

(f) [HNO3] = 0.0046 mol / 0.0220 L = 0.209 M

(g): pH = pKa + log[NaNO2(aq)] / [HNO2]

=> pH = 3.14 + log(0.091M / 0.209M) = 2.78 (answer)

Note: Please conform the pKa value is 3.14 or not


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