Calculate the pH of a solution obtained by mixing 50 mL of 0.75 M CH3COOH with 15 mL of 1.6 M NaOH. (Ka of CH3COOH is 1.8 x 10-5)
To calculate the pH of the solution, we need to determine the concentration of the CH3COOH and NaOH after they neutralize each other. We can use the concept of stoichiometry to find the limiting reactant and the remaining excess reactant.
First, let's determine the amount of moles for each reactant:
Moles of CH3COOH = volume (in L) x concentration (in M)
= (50 mL / 1000 mL/L) x 0.75 M
= 0.0375 moles
Moles of NaOH = volume (in L) x concentration (in M)
= (15 mL / 1000 mL/L) x 1.6 M
= 0.024 moles
Next, let's compare the moles of CH3COOH and NaOH to find the limiting reactant.
According to the balanced equation for the reaction between CH3COOH and NaOH:
CH3COOH + NaOH → CH3COONa + H2O
The ratio of moles of CH3COOH to NaOH is 1:1. Therefore, the limiting reactant is the one with the smaller number of moles, which is NaOH in this case.
Since the amount of NaOH is consumed fully, we can use the remaining moles of CH3COOH to calculate the molarity of CH3COOH in the final solution.
Final moles of CH3COOH = initial moles of CH3COOH - moles of NaOH
= 0.0375 moles - 0.024 moles
= 0.0135 moles
Total volume of the final solution = volume of CH3COOH + volume of NaOH
= 50 mL + 15 mL
= 65 mL
Convert the total volume to liters:
Total volume = 65 mL / 1000 mL/L
= 0.065 L
Finally, we can use the formula for calculating pH of a weak acid solution:
pH = -log10([H+])
Since CH3COOH is a weak acid, we can assume that it does not fully ionize. Therefore, the concentration of [H+] can be approximated as the concentration of CH3COOH.
[H+] = moles of CH3COOH / total volume (in liters)
= 0.0135 moles / 0.065 L
≈ 0.2077 M
Now, we can calculate the pH:
pH = -log10(0.2077)
≈ 0.68
Therefore, the pH of the solution obtained by mixing 50 mL of 0.75 M CH3COOH with 15 mL of 1.6 M NaOH is approximately 0.68.