Problem 1c: Suppose 75.00 mL of 0.310 M HBr and 93.00 mL 0.250 M KOH are mixed together.A. Which reactant is the limiting reactant?B. Build the reaction table to describe the species that remain at the end of the reaction.RI (moles)C (moles)E (moles)C. Which species remain in solution after the addition of 93.00 mL of KOH?D. What do you think the pH would be at this point: acidic, basic or neutral?

Answered: Problem 1c: Suppose 75.00 mL of 0.310 M…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Problem 1c: Suppose 75.00 mL of 0.310 M HBr and 93.00 mL 0.250 M KOH are mixed together.
A. Which reactant is the limiting reactant?
B. Build the reaction table to describe the species that remain at the end of the reaction.
R
I (moles)
C (moles)
E (moles)
C. Which species remain in solution after the addition of 93.00 mL of KOH?
D. What do you think the pH would be at this point: acidic, basic or neutral?

Expert Solution

Step 1

The balanced equation for the neutralization reaction between $HBr and KOH$ is as follows.

$HBr + KOH \to KBr + H_{2} O$

It is also a double replacement reaction where cations or anions of $HBr and KOH$ got exchanged.

From the equation we know that we require $1 mol HBr for reacting with 1 mol KOH .$

Step 2

A) The species that got consumed completely in a reaction is called the limiting reagent of that reaction.

The no of moles of $HBr and KOH$ available with us can be calculated as follows.

$\begin{array}{rcl} Molarity (M) & = & \frac{moles of solute (mol)}{volume of solution (L)} \\ moles of solute & = & volume of solution (L) \times Molarity (M) \\ moles of HBr & = & 75 \times 10^{- 3} L \times 0.310 \frac{mol}{L} \\ = & 0.02325 mol \\ moles of KOH & = & 93 \times 10^{- 3} L \times 0.250 \frac{mol}{L} \\ = & 0.02325 \end{array}$

That means both reactants are consumed completely in the given reaction.

Step 3

B) In the mixture we have $0.02325 mol HBr and KOH$ . So after reaction we will get $0.02325 mol KBr and H_{2} O .$

Now we can built the reaction table to describe the species that remain at the end of the reaction.

$R$	$HBr$	$KOH$	$KBr$	$H_{2} O$
$I (moles)$	$0.02325$	$0.02325$	$0$	$0$
$C (moles)$	$- 0.02325$	$- 0.02325$	$+ 0.02325$	$+ 0.02325$
$E (moles)$	$0$	$0$	$0.02325$	$0.02325$

Hence the species that remain at the end of the reaction are $KOH and H_{2} O$ .

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