chemistry question, titrations

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a 30.0mL sample of .220M sulfuric acid is titrated with potassium hydroxide. if it takes 26.4mL of base solution to reach the endpoint, what is the molarity of the base solution?

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Explanation & Answer

Firstly, we can write down the chemical balanced equation as follow:

H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)

Conversion of units:
30.0 mL = 0.03 L
26.4 mL = 0.0264 L

The moles of sulfuric acid used can be calculated by:

nacid = Macid x Vacid = (0.220 M) x (0.03 L) = 6.6 x 10-3 mol

According to the balanced equation, since 1 mol of H2SO4 is reacted with 2 mol of KOH, we have that the moles of potassium hydroxide is equal to

nbase = 2 x 6.6 x 10-3 mol = 1.32 x 10-2 mol

Finally, the molarity of the 26.4mL base solution is determined by:

Mbase = nbase/ Vbase = ( 1.32 x 10-2 mol) / (0.0264 L) = 0.500 M

Answer: 0.500 M




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