Answer:
The pOH = 1.83
Explanation:
Step 1: Data given
volume of the sample = 25.0 mL
Molarity of hydrocyanic acid = 0.150 M
Molarity of NaOH = 0.150 M
Ka of hydrocyanic acid = 4.9 * 10^-10
Step 2: The balanced equation
HCN + NaOH → NaCN + H2O
Step 3: Calculate the number of moles hydrocyanic acid (HCN)
Moles HCN = molarity * volume
Moles HCN = 0.150 M * 0.0250 L
Moles HCN = 0.00375 moles
Step 3: Calculate moles NaOH
Moles NaOH = 0.150 M * 0.0305 L
Moles NaOH = 0.004575 moles
Step 4: Calculate the limiting reactant
0.00375 moles HCN will react with 0.004575 moles NaOH
HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH
Step 5: Calculate molarity of NaOH
Molarity NaOH = moles NaOH / volume
Molarity NaOH = 0.000825 moles / 0.0555 L
Molarity NaOH = 0.0149 M
Step 6: Calculate pOH
pOH = -log [OH-]
pOH = -log (0.0149)
pOH = 1.83
The pOH = 1.83
