What is the pH of 125 mL of a 1.59 M solution of a monoprotic weak acid, HA (Ka=9.53×10–5), after 29 mL of 0.37 M NaOH is added?
I have no Idea how to start this problem and the homework is due tonight, could someone please help me in calculating this? Thanks
Of course! I'll be happy to help you solve this problem.
To find the pH of the solution, we need to consider the reaction between the weak acid, HA, and the strong base, NaOH. In this case, since HA is a monoprotic weak acid, it donates one proton (H+) to the hydroxide ion (OH-) from NaOH to form water (H2O).
To determine the pH, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction between HA and NaOH:
HA + NaOH → H2O + NaA
Step 2: Calculate the initial moles of HA before the addition of NaOH.
Given the volume of the HA solution is 125 mL and the concentration is 1.59 M, we can calculate the initial moles of HA using the formula:
moles = concentration × volume (in liters)
So, moles of HA = 1.59 M × (125 mL / 1000 mL) = 0.19875 moles
Step 3: Determine the limiting reagent.
To determine the limiting reagent, we compare the moles of HA with the moles of NaOH. Since the balanced equation shows a 1:1 stoichiometric ratio between HA and NaOH, the limiting reagent will be the one with fewer moles. In this case, NaOH is the limiting reagent.
Step 4: Calculate the moles of excess NaOH.
Given the volume of NaOH added is 29 mL and the concentration is 0.37 M, we can calculate the moles of NaOH added using the formula:
moles = concentration × volume (in liters)
So, moles of NaOH added = 0.37 M × (29 mL / 1000 mL) = 0.01073 moles
Since NaOH is the limiting reagent, the moles of NaOH added will react completely with the moles of HA.
Step 5: Determine the moles of HA that reacted.
Since the balanced equation shows a 1:1 stoichiometric ratio between HA and NaOH, the moles of HA that reacted will be equal to the moles of NaOH added, which is 0.01073 moles.
Step 6: Calculate the remaining moles of HA.
The initial moles of HA (0.19875 moles) minus the moles of HA that reacted (0.01073 moles) will give us the remaining moles of HA.
So, remaining moles of HA = 0.19875 moles - 0.01073 moles = 0.18802 moles
Step 7: Calculate the concentration of the remaining HA.
Given the volume of the solution is 125 mL minus the volume of NaOH added (29 mL), we can calculate the concentration of the remaining HA using the formula:
concentration = moles / volume (in liters)
So, concentration of remaining HA = 0.18802 moles / (125 mL - 29 mL) / 1000 mL = 1.663 M
Step 8: Calculate the pH.
To calculate the pH, we need the concentration of H+ ions produced by the remaining HA. Since HA is a weak acid, we can consider it to be partially ionized.
Using the equation for Ka (acid dissociation constant) for the weak acid, we can calculate the concentration of H+ ions using the formula:
Ka = [H+][A-] / [HA]
Given the Ka value of 9.53×10–5 and the concentration of remaining HA as 1.663 M, we can rearrange the equation to solve for the concentration of H+ ions:
[H+] = sqrt(Ka × [HA])
So, [H+] = sqrt(9.53×10–5 × 1.663) = 7.271×10–3 M
Remember that pH is calculated as the negative logarithm of the concentration of H+ ions:
pH = -log[H+]
So, pH = -log(7.271×10–3) = 2.14
Therefore, the pH of the solution after adding 29 mL of 0.37 M NaOH is approximately 2.14.
I hope this explanation helps you understand how to calculate the pH in this type of problem. Let me know if you have any further questions!