Q: 26.30mL of 0.1105 mol/L H3PO4 solution is neutralized by 139.50 mL of KOH.
Calculate the mass of solid KOH required to prepare 500.0 mL of this solution.
In the previous part of this question, it asked the concentration of KOH and I worked it out to be concentration of KOH is: 0.062497 mol/L
I don't know how to do that mass calculation.
I checked your calculation for (KOH) and it is correct at 0.06250 (rounded to 4 s.f.).
Then how many mols do you for 500 mL of the solution. If it is 0.06250 mols/L then for 1/2 L you will need 1/2 that = ?
Then grams KOH = mols KOH x molar mass KOH = ?
To calculate the mass of solid KOH required to prepare 500.0 mL of the solution, we need to use the balanced chemical equation and stoichiometry.
First, let's review the balanced chemical equation for the reaction between H3PO4 (phosphoric acid) and KOH (potassium hydroxide):
H3PO4 + 3KOH -> K3PO4 + 3H2O
From the balanced equation, we can see that the mole ratio between H3PO4 and KOH is 1:3. This means that for every 1 mole of H3PO4, we need 3 moles of KOH to react.
In the previous part of the question, you calculated the concentration of KOH to be 0.062497 mol/L. Now, we can use this information to find the number of moles of KOH.
Number of moles of KOH = concentration of KOH × volume of KOH used
= 0.062497 mol/L × 0.13950 L
≈ 0.008717 moles
Next, we can use the mole ratio between H3PO4 and KOH to determine the number of moles of H3PO4.
Number of moles of H3PO4 = number of moles of KOH × (1 mole H3PO4 / 3 moles KOH)
≈ 0.008717 moles × (1/3)
≈ 0.002906 moles
Now, let's calculate the molar mass of KOH. The molar mass of K (potassium) is approximately 39.10 g/mol, the molar mass of O (oxygen) is approximately 16.00 g/mol, and the molar mass of H (hydrogen) is approximately 1.01 g/mol.
Molar mass of KOH = (molar mass of K) + (molar mass of O) + (molar mass of H)
= (39.10 g/mol) + (16.00 g/mol) + (1.01 g/mol)
≈ 56.11 g/mol
Now, we can calculate the mass of solid KOH required to prepare 500.0 mL of the solution.
Mass of solid KOH = number of moles of KOH × molar mass of KOH
≈ 0.008717 moles × 56.11 g/mol
≈ 0.488 g (rounded to three decimal places)
Therefore, to prepare 500.0 mL of the solution, you would need approximately 0.488 grams of solid KOH.