Enter your answer in the provided box. The volume of a sample of pure HCl gas wa
ID: 508432 • Letter: E
Question
Enter your answer in the provided box.The volume of a sample of pure HCl gas was 255 mL at 25°C and 181 mmHg. It was completely dissolved in about 60 mL of water and titrated with an NaOH solution; 26.1 mL of the NaOH solution was required to neutralize the HCl. Calculate the molarity of the NaOH solution.
Enter your answer in the provided box.
The volume of a sample of pure HCl gas was 255 mL at 25°C and 181 mmHg. It was completely dissolved in about 60 mL of water and titrated with an NaOH solution; 26.1 mL of the NaOH solution was required to neutralize the HCl. Calculate the molarity of the NaOH solution.
Explanation / Answer
PV = nRT
moles of HCl = n = PV / RT
P = 181 mmHg = 0.238 atm
V = 255 ml = 0.255 L
T = 25 + 273.5 = 298.5 K
R = 0.082 atm.L/mol.K
moles of HCl = n = PV / RT
= 0.238 x 0.255 / (0.082 x 298.5)
= 0.00248 mol
to neutralize HCl, moles of NaOH = moles of HCl = 0.00248 mol
molarity of NaOH = moles of NaOH / volume of NaOH
= 0.00248 mol / (26.1/1000) L
= 0.0950 M
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