What is the pH of a solution formed by mixing 125.0 mL of 0.0250 M HCl with 75.0 mL of 0.0500 M NaOH?
Solution 109P :
Volume of HCl = 125.0 mL
Molarity of HCl = 0.0250 M
Volume of NaOH = 75.0 mL
Molarity of NaOH = 0.0500 M
pH of solution formed by mixing HCl and NaOH= ?
Here, HCl is a strong acid and NaOH is a strong base.
Let’s write the balanced chemical equation for it :
HCl(aq) + NaOH(aq) → NaCl + H2O
First, let’s find the number of moles of HCl and NaOH :
We have the molarity and volume of HCl and NaOH, using this we will calculate the number of moles using the molarity formula :
The formula for calculating the number of moles will be :
Number of moles = Molarity Volume(in L)
Volume of HCl and NaOH is in mL, so let’s convert to L :
1 L = 1000 mL
Therefore, 125 mL HCl in L will be :
= 125 mL
= 0.125 L
75 mL NaOH in L will be :
= 75 mL
= 0.075 L
Hence, number of moles in HCl is :
= 0.0250 M 0.125 L
= 0.003125 moles.
Number of moles in NaOH is :
= 0.0500 M0.075 L
= 0.00375 moles.
Now, if we take the difference between [H3O] and [OH] :
= 0.00375 - 0.003125
= 0.000625 moles
This tells that there are more [OH] in solution. Hence, it is a basic solution and there are 0.000625 moles excess in the solution