Answer:
pH = 0.86
To solve this problem, we need to use the balanced chemical equation for the reaction between HNO3 and NaOH:
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
From the equation, we know that 1 mole of HNO3 reacts with 1 mole of NaOH in a 1:1 ratio.
First, we need to calculate the number of moles of HNO3 present in the initial solution:
moles of HNO3 = volume of HNO3 x concentration of HNO3
moles of HNO3 = 0.02284 L x 0.26 mol/L
moles of HNO3 = 0.0059344 mol
Next, we need to calculate the number of moles of NaOH added to the solution:
moles of NaOH = volume of NaOH x concentration of NaOH
moles of NaOH = 0.01172 L x 0.10 mol/L
moles of NaOH = 0.001172 mol
Since we added less moles of NaOH than moles of HNO3 present in the initial solution, the HNO3 will be the limiting reactant and all the NaOH will be used up in the reaction. Therefore, we can calculate the number of moles of HNO3 remaining after the reaction:
moles of HNO3 remaining = moles of HNO3 - moles of NaOH
moles of HNO3 remaining = 0.0059344 mol - 0.001172 mol
moles of HNO3 remaining = 0.0047624 mol
To calculate the volume of the solution after the addition of NaOH, we add the volumes of the initial HNO3 solution and the NaOH solution:
volume of solution = volume of HNO3 + volume of NaOH
volume of solution = 0.02284 L + 0.01172 L
volume of solution = 0.03456 L
Finally, we can use the definition of pH to calculate the pH of the remaining HNO3 solution:
pH = -log[H3O+]
[H3O+] = moles of HNO3 remaining / volume of solution
[H3O+] = 0.0047624 mol / 0.03456 L
[H3O+] = 0.1379 M
pH = -log(0.1379)
pH = 0.86
Therefore, the pH of the remaining HNO3 solution after 11.72 mL of 0.10 M NaOH have been added is approximately 0.86.