What is the volume of 0.129 M HCl(aq) re-
quired to neutralize all of the hydroxide ions in 46 mL of 0.216 M KOH(aq)?
Answer in units of mL.
Answer: 77.0232
What is the molarity of Cl− ions at the stoichiometric point?
Answer in units of mol/L.
What is the pH of the solution after the addition of 49.78 mL of 0.129 M HCl(aq) to 46 mL of 0.216 M KOH(aq)?
Answer
To find the volume of HCl(aq) required to neutralize all of the hydroxide ions in the KOH(aq) solution, you need to determine the stoichiometry of the reaction between HCl and KOH. The balanced chemical equation for this reaction is:
HCl + KOH -> KCl + H2O
From the equation, you can see that one mole of HCl reacts with one mole of KOH to produce one mole of water.
Step 1: Calculate the number of moles of KOH in the 46 mL solution.
To find the number of moles, use the formula:
moles = concentration (M) × volume (L)
moles of KOH = 0.216 M × 46 mL (convert to L) = 0.216 M × 0.046 L = 0.009936 moles
Step 2: Calculate the volume of HCl solution required to neutralize all of the hydroxide ions.
Since the reaction is 1:1, the moles of HCl required will be the same as the moles of KOH.
moles of HCl = 0.009936 moles
To find the volume of HCl solution, use the formula:
volume (L) = moles / concentration (M)
volume of HCl = 0.009936 moles / 0.129 M = 0.0770232 L
Finally, convert the volume from liters to milliliters:
volume of HCl = 0.0770232 L × 1000 mL/L ≈ 77.0232 mL
Therefore, the volume of 0.129 M HCl(aq) required to neutralize all of the hydroxide ions in 46 mL of 0.216 M KOH(aq) is approximately 77.0232 mL.
To find the molarity of Cl− ions at the stoichiometric point, we need to calculate the number of moles of Cl− ions formed when all of the hydroxide ions are neutralized.
From the balanced equation:
HCl + KOH -> KCl + H2O
we can see that one mole of HCl reacts with one mole of KCl, and each mole of KCl produces one mole of Cl− ions.
Since both HCl and KOH are monoprotic acids and bases, the stoichiometry is 1:1.
Using the number of moles of KOH calculated before (0.009936 moles), the number of moles of Cl− ions formed will also be 0.009936 moles.
To find the molarity, divide the moles of Cl− ions by the volume of the solution in liters:
molarity of Cl− ions = moles of Cl− ions / volume of solution (L)
The volume of the solution is the sum of the volumes of HCl and KOH used.
With the volume of HCl determined to be 77.0232 mL (convert to L) and the volume of KOH as 46 mL (convert to L):
volume of solution = 77.0232 mL + 46 mL = 123.0232 mL = 0.1230232 L
molarity of Cl− ions = 0.009936 moles / 0.1230232 L
Therefore, the molarity of Cl− ions at the stoichiometric point is approximately 0.0807 mol/L.
To find the pH of the solution after the addition of 49.78 mL of 0.129 M HCl(aq) to 46 mL of 0.216 M KOH(aq), we need to consider the reaction that occurs between the two solutions.
HCl, being a strong acid, dissociates completely in water to produce H+ ions. Similarly, KOH, being a strong base, dissociates completely in water to produce OH− ions.
The reaction between HCl and KOH can be represented as:
HCl(aq) + KOH(aq) -> KCl(aq) + H2O(l)
Since HCl is a strong acid and KOH is a strong base, the reaction goes to completion with no excess reactant remaining.
Step 1: Calculate the number of moles of HCl in the 49.78 mL solution.
moles of HCl = concentration (M) × volume (L)
moles of HCl = 0.129 M × 49.78 mL (convert to L) = 0.129 M × 0.04978 L = 0.00642562 moles
Step 2: Calculate the total moles of H+ ions and OH− ions in solution.
The moles of H+ ions will be equal to the moles of HCl added, while the moles of OH− ions will be equal to the moles of KOH in the initial solution.
moles of H+ ions = 0.00642562 moles
moles of OH− ions = 0.009936 moles (calculated previously)
Step 3: Calculate the concentration of H+ ions and OH− ions.
The concentration of H+ ions is given by the formula:
concentration (M) = moles / volume (L)
concentration of H+ ions = 0.00642562 moles / (46 mL + 49.78 mL) (convert to L)
concentration of OH− ions = 0.009936 moles / (46 mL + 49.78 mL) (convert to L)
Step 4: Calculate the pOH and pH of the solution.
pOH = −log10(concentration of OH− ions)
pH = 14 − pOH
Therefore, using the calculated concentrations, determine the pOH and pH values.