A4.48 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?

27.53 mL.

Explanation:

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.

(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.

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error