Question

Solution A is 0.1mol/dm² HCI and solution B contains 5g of Impure NaOH in one liter. The average volume of acid

used to neutralize 25cm^{3} the base was determined to be 20.50~cm^{3}

Answer 1

To find the concentration of solution B, we need to determine the number of moles of NaOH in the 5g sample.

The molar mass of NaOH is 23 + 16 + 1 = 40 g/mol.
Therefore, the number of moles of NaOH is 5g / 40 g/mol = 0.125 mol.

Since the volume of solution B is 1 liter, the concentration of NaOH in solution B is 0.125 mol/1 L = 0.125 mol/L.

Now we can calculate the number of moles of HCl used to neutralize the NaOH in 25 cm^3 of solution B.

The average volume of HCl used to neutralize the base is 20.50 cm^3.
Since the concentration of solution A is 0.1 mol/dm^3, the number of moles of HCl used is (20.50 cm^3 / 1000 cm^3/dm^3) * 0.1 mol/dm^3 = 0.00205 mol.

Since the reaction between HCl and NaOH is 1:1, we know that 0.00205 mol of HCl neutralizes 0.00205 mol of NaOH.

To find the volume of HCl that would neutralize 0.00205 mol of NaOH, we can use the proportion:

(0.125 mol NaOH / 0.00205 mol NaOH) * 25 cm^3 = 1524.39 cm^3

Therefore, the average volume of HCl needed to neutralize 25 cm^3 of solution B is approximately 1524.39 cm^3.