A.) An aqueous solution contains 0.458 M ammonia . Calculate the pH of the solut

Question

A.) An aqueous solution contains 0.458 Mammonia.

Calculate the pH of the solution after the addition of 3.41×10-2 moles of nitric acid to 155 mL of this solution.
(Assume that the volume does not change upon adding nitric acid.)

pH =

B.) An aqueous solution contains 0.377 Mhydrocyanic acid.

Calculate the pH of the solution after the addition of 4.97×10-2 moles ofpotassium hydroxide to 255 mL of this solution.
(Assume that the volume does not change upon adding potassium hydroxide)

pH =

C.) An aqueous solution contains 0.338 Mhydrofluoric acid.

Calculate the pH of the solution after the addition of 1.95×10-2 moles of sodium hydroxide to 125 mL of this solution.
(Assume that the volume does not change upon adding sodium hydroxide)

pH =

Explanation / Answer

A)
How many moles of base are contained in 155 ml of the ammonia solution?

nNH3 = 0.458 M . 155 ml . (1 L/1000 ml)
nNH3 = 0.07099 mol

nHNO3 = 3.41x10-2 mol

Chemical equation for the reaction.
NH3 + HNO3 --> NH4+ + NO3-

nNH3 > nHNO3
The nitric acid is the limiting reactant, so are consumed 3.41x10-2 mol of NH3 to generate 3.41x10-2 mol NH4+.

After the chemical reaction:
nNH3 = 0.07099 mol - 3.41x10-2 mol
nNH3 = 3.689x10-2 mol

nNH4+ =3.41x10-2 mol

As the acid react with the ammonia, a buffer is formed.
You must use the Henderson-Hasselbalch equation to determine the pH or pOH of the solution.

pOH = pKb + log([HB+]/[B])
pOH = pKb + log(nHB+/nB)
*equivalent mathematical equations

pKb = 4.75 (for ammonia in water)
nHB+, moles of acid (NH4+)
nB, moles of base (NH3)

pOH = 4.75 + log(3.41x10-2 mol/3.689x10-2 mol)
pOH = 4.72

pH + pOH = 14
pH = 14 - pOH
pH = 14 - 4.72
pH = 9.28

______________________________

B)
How many moles of acid are contained in 255 ml of the hydrocyanic solution?

nHCN = 0.377 M . 255 ml . (1 L/1000 ml)
nHCN = 0.096135 mol

nKOH = 4.97x10-2 mol

Chemical equation for the reaction.
HCN + KOH --> KCN + H2O

nHCN > nKOH
The potassium hydroxide is the limiting reactant, so are consumed 4.97x10-2 mol of HCN to generate 4.97x10-2 mol CN-.

After the chemical reaction:
nHCN = 0.096135 mol - 4.97x10-2 mol
nHCN = 4.6435x10-2 mol

nCN- =4.97x10-2 mol

As the base react with the hydrocyanic acid, a buffer is formed.
You must use the Henderson-Hasselbalch equation to determine the pH or pOH of the solution.

pH = pKa + log([A-]/[HA])
pH = pKa + log(nA-/nHA)
*equivalent mathematical equations

pKa = 9.2 (for hydrocyanic acid in water)
nHA, moles of acid (HCN)
nA-, moles of base (CN-)

pH = 9.20 + log(4.97x10-2 mol/4.6435x10-2 mol)
pH = 9.23

______________________________

C)
How many moles of acid are contained in 125 ml of the hydrofluoric solution?

nHF = 0.338 M . 125 ml . (1 L/1000 ml)
nHF = 0.04225 mol

nNaOH = 1.95x10-2 mol

Chemical equation for the reaction.
HF + NaOH --> NaF + H2O

nHF > nNaOH
The sodium hydroxide is the limiting reactant, so are consumed 1.95x10-2 mol of HF to generate 1.95x10-2 mol F-.

After the chemical reaction:
nHF = 0.04225 mol - 1.95x10-2 mol
nHF = 2.275x10-2 mol

nF- =1.95x10-2 mol

As the base react with the hydrofluoric acid, a buffer is formed.
You must use the Henderson-Hasselbalch equation to determine the pH or pOH of the solution.

pH = pKa + log([A-]/[HA])
pH = pKa + log(nA-/nHA)
*equivalent mathematical equations

pKa = 3.17 (for hydrofluoric acid in water)
nHA, moles of acid (HF)
nA-, moles of base (F-)

pH = 3.17 + log(1.95x10-2 mol /2.275x10-2 mol)
pH = 3.10

______________________________

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