The molarity of an aqueous solution of hydroiodic acid, HI, is determined by titration with a 0.145 M potassium hydroxide, KOH, solution. If 45.7 mL of potassium hydroxide solution are required to neutralize 50.0 mL of the acid, what is the molarity of the hydroiodic acid solution?

Question

Chemistry

Avery Osborn

1 March, 20:48

The molarity of an aqueous solution of hydroiodic acid, HI, is determined by titration with a 0.145 M potassium hydroxide, KOH, solution. If 45.7 mL of potassium hydroxide solution are required to neutralize 50.0 mL of the acid, what is the molarity of the hydroiodic acid solution?

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Katelyn Maxwell · Answer 1

0.133 M

Explanation:

The volume of the solution is given, so in order to find concentration, the number of moles must be found, since C = n/V.

The balanced reaction equation is:

HI + KOH ⇒ H₂O + KI

Thus, the moles of KOH added to neutralize all of the HI will be equal to the moles of HI that must have been present.

The amount of KOH that was added is calculated as follows.

n = CV = (0.145 mol/L) (45.7 mL) = 6.6265 mmol KOH = 6.6265 mmol HI

Since HI and KOH are related in a 1:1 molar ratio, the same amount of HI must have been present.

Finally, the concentration of HI is calculated:

C = n/V = (6.6265 mmol) / (50.0 mL) = 0.133 mol/L = 0.133 M