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:
H2SO4(aq) + 2KOH(aq) → K2SO4(aq)
nacid = Macid x Vacid
= (0.220 M) x (0.03 L) = 6.6 x 10-3
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
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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