What is the pH of a solution formed by mixing 175.0 mL of 0.0880 M HI with 125.0 mL of 0.0570 M KOH?
Problem 110P :
Volume of HI = 175.0 mL
Molarity of HI = 0.0880 M
Volume of KOH = 125.0 mL
Molarity of KOH = 0.0570 M
pH of solution formed by mixing HI and KOH= ?
Here, HI is a strong acid and KOH is a strong base.
Let’s write the balanced chemical equation for it :
HI(aq) + KOH(aq) → KI + H2O
First, let’s find the number of moles of HI and KOH :
We have the molarity and volume of HI and KOH, 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 HI and KOH is in mL, so let’s convert to L :
1 L = 1000 mL
Therefore, 175 mL HI in L will be :
= 175 mL
= 0.175 L
125 mL KOH in L will be :
= 125 mL
= 0.125 L
Hence, number of moles in HI is :
= 0.0880 M 0.175 L
= 0.0154 moles.
Number of moles in KOH is :
= 0.0570 M0.125 L
= 0.007125 moles.
Now, if we take the difference between [H3O] and [OH] :
= 0.0154 - 0.007125
= 0.008275 moles
This tells that there are more [H+] ions in solution. Hence, it is a acidic solution and there are 0.008275 moles excess in the solution