What is the final pH of a solution made by mixing 36.4mL of 0.25mol/L HCl(aq) with 20.0mL of 0.051mol/L NaOH(aq)?
To determine the final pH of the solution, we need to calculate the amount of H+ and OH- ions in the solution after mixing.
The balanced chemical equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
According to the equation, the stoichiometric ratio between HCl and NaOH is 1:1. This means that in order for the reaction to go to completion, both HCl and NaOH must be completely consumed.
First, let's calculate the number of moles of HCl:
Moles of HCl = volume of HCl solution (in L) × concentration of HCl
= 36.4 mL × (1 L / 1000 mL) × 0.25 mol/L
= 0.0091 mol
Similarly, let's calculate the number of moles of NaOH:
Moles of NaOH = volume of NaOH solution (in L) × concentration of NaOH
= 20.0 mL × (1 L / 1000 mL) × 0.051 mol/L
= 0.0102 mol
Since the stoichiometric ratio between HCl and NaOH is 1:1, the number of moles for HCl and NaOH consumed will be equal.
However, since the volume of HCl is less than the volume of NaOH, there will be some leftover NaOH. To calculate the excess moles of NaOH:
Excess moles of NaOH = moles of NaOH - moles of HCl
= 0.0102 mol - 0.0091 mol
= 0.0011 mol
Now, let's calculate the concentration of the remaining OH- ions in the solution:
Concentration of OH- = excess moles of NaOH / total volume of solution (in L)
= 0.0011 mol / (36.4 mL + 20.0 mL) × (1 L / 1000 mL)
= 0.0011 mol / 0.0564 L
= 0.0195 mol/L
Since we have excess OH- ions and no H+ ions remaining, the final solution will be basic.
To calculate the pH, we can use the pOH formula:
pOH = -log10(concentration of OH-)
= -log10(0.0195)
≈ 1.71
Finally, we can calculate the pH using the equation:
pH + pOH = 14
pH = 14 - pOH
= 14 - 1.71
≈ 12.29
Therefore, the final pH of the solution is approximately 12.29.