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OlgaM077 [116]
4 years ago
8

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 t

his solution is titrated with a 0.227 M Sr(OH)2 solution. What volume of standard solution is required to reach the equivalence point?
Chemistry
1 answer:
serious [3.7K]4 years ago
8 0

Answer:

27.53 mL.

Explanation:

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

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).

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

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

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

<em>(MV)HCl = (xMV) Sr(OH)₂</em>

∴ 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.

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