Question

How many milliliters of 0.550 M hydriodic acid are needed to react with 25.00 mL of 0.217 M CsOH?

Answer 1

Balanced equation is

HI + CsOH —> CsI + H2O
(Double replacement)

You can find the moles of CsOH by multiplying liters and molarity together. The formula is.

M = n / V
M is molarity, n is moles and is volume in liters

First change 25.00 mL to L by dividing 1000
• 0.025 L
Now multiply molarity and liters to get moles of CsOH

0.025 L x 0.217 M = 0.0054 moles CsOH

since the balanced equation is a one to one ratio, all compounds will be 0.0054 moles, in this case HI acid has 0.0054 moles.

Now you can find volume in liters by using the moles divided by the molarity of HI.

0.0054 moles HI / 0.550 M HI =
0.0098 L of HI

Now just change liters back to milliliters by multiplying by 1000

0.0098 x 1000 = 9.8 mL of HI needed

Answer 2

The acids react with base and the reaction is known as neutralization

One mole of hydronium ion reacts with one mole of hydroxide ion

Here

a) HI is a monoprotic acid : one mole of HI will give one mole of hydronium ion

b) CsOH is a monohydroxide base : one mole of CsOH will give one mole of hydroxide ion

So one mole of HI will react with one mole of CsOH

The moles of CsOH taken will be calculated from molarity and volume as

$Moles = Molarity X volume (L)$

Thus moles of CsOH are

$Moles of CsOH = 0.217 X \frac{25}{1000} = 0.005425$

moles of HI required is 0.005425 mol

volume of HI required will be

$Volume of HI=\frac{moles}{molarity} = \frac{0.005425}{0.550} = 0.00986 L = 9.86 mL$

Volume of HI required is 9.86 mL