A 50.00-mL solution of 0.0350 M hydrocyanic acid (K_a = 6.2 $\times$10^-¹⁰)is titrated with a 0.0333 M solution of sodium hydroxide as the titrant.What is the pH of the acid solution after 15.00 mL of titrant have been added? (K_w = 1.00 $\times$10^-¹⁴)

Question

Quizplus · Accepted Answer

First, let's write the balanced chemical equation for the reaction between hydrocyanic acid and sodium hydroxide:

HCN (aq) + NaOH (aq) → NaCN (aq) + H2O (l)

This is a one-to-one reaction, which means that every mole of NaOH will react with one mole of HCN.

To determine the pH of the solution after 15.00 mL of titrant have been added, we need to calculate the number of moles of HCN that are left in the solution.

Before the titration: 
- Number of moles of HCN = molarity x volume = 0.0350 M x 0.0500 L = 0.00175 moles

After adding 15.00 mL of 0.0333 M NaOH: 
- Number of moles of NaOH = molarity x volume = 0.0333 M x 0.01500 L = 0.00050 moles 
- Number of moles of HCN left = 0.00175 moles - 0.00050 moles = 0.00125 moles

Now, let's use the Henderson-Hasselbalch equation to calculate the pH of the solution:

pH = pKa + log ([A-]/[HA])

where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base (in this case, CN-), and [HA] is the concentration of the acid (in this case, HCN).

The pKa of HCN is given as 6.2. The concentration of CN- can be calculated from the number of moles of NaOH that were added:

- Number of moles of CN- = number of moles of NaOH = 0.00050 moles 
- Volume of final solution = volume of acid + volume of titrant = 50.00 mL + 15.00 mL = 65.00 mL = 0.0650 L 
- Concentration of CN- = number of moles / volume = 0.00050 moles / 0.0650 L = 0.00769 M

Substituting these values into the Henderson-Hasselbalch equation:

pH = 6.2 + log (0.00769 M / 0.00125 M) = 8.81

Therefore, the pH of the acid solution after 15.00 mL of titrant have been added is 8.81, which corresponds to choice C.

A 5000-ML Solution of 0 ×\times× 10-10)is Titrated with a 0 ×\times×

A 5000-ML Solution of 0 $\times$ 10^-¹⁰)is Titrated with a 0 $\times$