CHM1 15 Example 2 Collection

Since we are given the molarity of the strong acid and strong base as well as the volume of the base, we are able to find the volume of the acid. The equation of the reaction is as follows:

`HI(aq)+KOH(aq) -> H_2 O(l)+KI(aq)`

We see that the mole ratio necessary for HI to neutralize KOH is 1:1; therefore, we need the moles of HI to be equal to the KOH present in the solution.

To find the number of moles of KOH we multiply the molarity of KOH with the volume of KOH, notice how the liter unit cancels out:

As the moles of KOH = moles of HI at the equivalence point, we have 4.2 moles of HI.

To find the volume of the solution of HI, we use the molarity of HI (3.4 M) and the fact that we have 4.2 moles of HI:

By dividing by 4.2 mol by 3.4 mol HI , we get:

We are left with X = 1.2 L. The answer is 1.2 L of 3.4 M HI required to reach the equivalence point with 2.1 L of 2.0 M KOH.

• Alternatively, as the required mole ratio of HI to KOH is 1:1, we can use the equation M₁V₁ = M₂V₂ to solve the problem:

(3.4 M)(V₁) = (2.1 L)(2.0 M)

V₁ = (2.1 L)(2.0 M) / (3.4 M) = 1.2