Answer :
Final answer:
The pH after adding 46.5 mL of base is approximately 8.83.
Explanation:
To calculate the pH after adding 46.5 mL of base, we need to determine the moles of acid and base present in the solution.
First, let's calculate the moles of hydrocyanic acid (HCN) in the initial 40.2 mL sample:
Volume of HCN solution = 40.2 mL = 0.0402 L
Moles of HCN = volume (L) x concentration (M) = 0.0402 L x 0.592 M = 0.0238 moles
Since the reaction between HCN and KOH is 1:1, the moles of hydroxide ions (OH-) in the 46.5 mL of base added is also 0.0238 moles.
Now, let's calculate the concentration of hydroxide ions (OH-) in the solution after adding the base:
Total volume of solution = initial volume of HCN solution + volume of base added = 40.2 mL + 46.5 mL = 86.7 mL = 0.0867 L
Concentration of OH- = moles of OH- / volume of solution = 0.0238 moles / 0.0867 L = 0.274 M
Since hydrocyanic acid is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-]/[HA])
The pKa value for hydrocyanic acid is known to be 9.21.
Since the reaction is 1:1, the concentration of the conjugate base ([A-]) is equal to the concentration of hydroxide ions (OH-), which is 0.274 M.
The concentration of the acid ([HA]) is the initial concentration of hydrocyanic acid, which is 0.592 M.
Plugging in the values into the Henderson-Hasselbalch equation:
pH = 9.21 + log(0.274/0.592) = 9.21 - 0.38 = 8.83
Therefore, the pH after adding 46.5 mL of base is approximately 8.83.
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