7.0 mL of 0.64 M HBr reacts with 38 mL of 0.39 M KOH.
1) What are the moles of acid before the reaction?
2) What are the moles of base before the reaction?
3) What is the limiting reagent?
4) How many moles of excess reagent are there after the reaction?
5) what is the concentration of the excess reagent after the reaction?
6) What is the pH of the solution?
never mind, I figured it out
To find the answers to these questions, we need to follow a step-by-step approach. Let's dive in:
1) To find the moles of acid before the reaction, we can use the equation: moles = concentration * volume.
Given:
Concentration of HBr = 0.64 M
Volume of HBr = 7.0 mL = 0.007 L
Using the equation, we can calculate the moles of acid:
Moles of HBr = 0.64 M * 0.007 L = 0.00448 moles
Therefore, there are 0.00448 moles of acid before the reaction.
2) To find the moles of base before the reaction, we can use the same equation: moles = concentration * volume.
Given:
Concentration of KOH = 0.39 M
Volume of KOH = 38 mL = 0.038 L
Using the equation, we can calculate the moles of base:
Moles of KOH = 0.39 M * 0.038 L = 0.01482 moles
Therefore, there are 0.01482 moles of base before the reaction.
3) To determine the limiting reagent, we need to compare the moles of acid and base using their stoichiometric ratio.
The balanced chemical equation for the reaction is:
HBr + KOH -> KBr + H2O
From the equation, we can see that the stoichiometric ratio is 1:1, meaning 1 mole of HBr reacts with 1 mole of KOH.
Given:
Moles of HBr = 0.00448 moles
Moles of KOH = 0.01482 moles
Comparing the moles of acid and base, we can see that the moles of HBr are smaller. Therefore, HBr is the limiting reagent.
4) To find the moles of excess reagent after the reaction, we need to determine the amount of the excess reagent that did not react. This can be done by subtracting the moles of the limiting reagent used from the moles of the excess reagent.
Since HBr is the limiting reagent, we need to determine how many moles of KOH reacted.
From the balanced chemical equation, we can see that 1 mole of KOH reacts with 1 mole of HBr. Therefore, the number of moles of KOH is the same as the moles of HBr used.
Moles of excess reagent (KOH) = Moles of KOH before the reaction - Moles of KOH used
Moles of excess reagent (KOH) = 0.01482 moles - 0.00448 moles = 0.01034 moles
Therefore, there are 0.01034 moles of excess KOH after the reaction.
5) To find the concentration of the excess reagent after the reaction, we need to calculate the new volume of the solution.
Given:
Volume of HBr = 7.0 mL = 0.007 L
Volume of KOH = 38 mL = 0.038 L
Moles of excess KOH = 0.01034 moles
Total volume of solution = Volume of HBr + Volume of KOH
Total volume of solution = 0.007 L + 0.038 L = 0.045 L
Using the equation, we can find the new concentration of the excess KOH:
Concentration of the excess KOH = Moles of excess KOH / Total volume of solution
Concentration of the excess KOH = 0.01034 moles / 0.045 L = 0.229 M
Therefore, the concentration of the excess KOH after the reaction is 0.229 M.
6) To determine the pH of the solution, we need to know the concentration of the resulting solution, as well as the dissociation constant (pKa) of the acid.
Since we have calculated the concentration of the excess KOH after the reaction (0.229 M), we can assume that the pH of the resulting solution will be close to neutral (pH 7), assuming the reaction goes to completion and produces water.
Note: The pKa of HBr is -9 (extremely acidic).
Therefore, the pH of the solution after the reaction is approximately 7