Question

A 4.48 L sample of HCl gas, measured at STP, is dissolved in enough water to produce 400.0 mL of solution. A 25.0 mL sample of this solution is titrated with a 0.227 M Sr(OH)2 solution. What volume of standard solution is required to reach the equivalence point?

Answer 1

Answer:

27.53 mL.

Explanation:

• Firstly, we need to calculate the no. of moles of HCl using the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.0 atm at STP).

V is the volume of the gas in L (4.48 L).

n is the no. of moles of the gas in mol.

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 273 K at STP).

∴ n = PV/RT = (1.0 atm)(4.48 L)/(0.0821 L.atm/mol.K)(273 K) = 0.1998 mol ≅ 0.12 mol.

∴ The concentration of 0.12 mol HCl in 400.0 mL water = n/V = (0.12 mol)/(0.4 L) = 0.499 ≅ 0.50 mol/L.

• At equivalence: the no. of millimoles of HCl = the no. of millimoles of Sr(OH)₂.

(MV)HCl = (xMV) Sr(OH)₂

∴ The volume of Sr(OH)₂ = (MV)HCl/(xM) Sr(OH)₂ = (25.0 mL)(0.50 mol/L)/(2)(0.227 M) = 27.53 mL.