Question

HI + RbOH →→ Rbl + H₂O

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In a titration experiment a 300 mL sample of hydrolodic acid (HI) 2.15 M is completely neutralized by 212 mL of an
aqueous solution of rubidium hydroxide (RbOH). What was the molarity of the initial RbOH?
6.45 M RbOH
(orange)
1.52 M RbOH
(magenta)
3.04 M RbOH
(aqua green)

Answer 1

The balanced chemical equation for the reaction between hydriodic acid (HI) and rubidium hydroxide (RbOH) is:

HI + RbOH → RbI + H2O

From the balanced equation, we can see that the stoichiometry of the reaction is 1:1, which means that 1 mole of HI reacts with 1 mole of RbOH.

Given that a 300 mL sample of HI with a concentration of 2.15 M is completely neutralized by 212 mL of RbOH, we can use the following equation to determine the molarity of RbOH:

Molarity of HI x Volume of HI = Molarity of RbOH x Volume of RbOH

2.15 M x 0.300 L = Molarity of RbOH x 0.212 L

Molarity of RbOH = (2.15 M x 0.300 L) / 0.212 L = 3.04 M

Therefore, the molarity of the initial RbOH solution was 3.04 M, option (aqua green).

Answer 2

The balanced chemical equation for the reaction between HI and RbOH is:

HI + RbOH → RbI + H2O

From the equation, we can see that 1 mole of HI reacts with 1 mole of RbOH to produce 1 mole of RbI and 1 mole of H2O.

To determine the molarity of the initial RbOH, we can use the following formula:

Molarity of RbOH = (moles of RbOH) / (volume of RbOH solution in liters)

First, let's calculate the number of moles of HI in the 300 mL sample:

moles of HI = Molarity x volume in liters

moles of HI = 2.15 mol/L x 0.3 L

moles of HI = 0.645 mol

Since 1 mole of HI reacts with 1 mole of RbOH, the number of moles of RbOH used in the titration is also 0.645 mol.

Now, we can use the volume and concentration of the RbOH solution to calculate its molarity:

Molarity of RbOH = (moles of RbOH) / (volume of RbOH solution in liters)

Molarity of RbOH = 0.645 mol / 0.212 L

Molarity of RbOH = 3.04 M

Therefore, the molarity of the initial RbOH solution was 3.04 M, which corresponds to the aqua green option.