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?

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

H_{2}SO_{4}(aq) + 2KOH(aq) → K_{2}SO_{4}(aq) + 2H_{2}O(l)

n_{acid} = M_{acid} x V_{acid} = (0.220 M) x (0.03 L) = 6.6 x 10^{-3} mol

According to the balanced equation, since 1 mol of H_{2}SO_{4} is reacted with 2 mol of KOH, we have that the moles of potassium hydroxide is equal to

n_{base} = 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:

M_{base} = n_{base}/ V_{base} = ( 1.32 x 10^{-2} mol) / (0.0264 L) = 0.500 M

Answer: 0.500 M_{}

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