# 0.365g of HCl gas was passed through 100cc of 0.2M NaOH solution.The pH of the resulting solution is

2
by kirubb

2014-05-30T19:32:36+05:30
Molarity of HCl = 0.365/(36.5*0.1)          (molarity = no. of mole / volume of solution in L.)
= 0.1M
HCl is a strong acid & NaOH is a strong base hence 0.1M of HCl neutralise 0.1M of NaOH and 0.1M of NaOH is remaining .
[OH-] = 0.1
[H+][OH-] = 10^-14
[H+]*10^-1 = 10^-14
[H+] = 10^-13
pH = -log[H+]
= 13              ( log10 = 1)

• Brainly User
2014-05-30T23:20:41+05:30
Molarity = no. of moles / volume in liters
On substituting the given values of NaOH in the above formula, we get                               No. of moles of NaOH =0.02

36.5 g of HCl ⇒ 1 mole
0.365 g of HCl ⇒ 0.01 mole

1 mole of NaOH reacts with 1 mole of HCl {From the equation, HCl + NaOH ⇒ NaCl + H₂O}
So, only 0.01 mole of NaOH reacts with 0.01 mole of HCl.

The Molarity of the un reacted 0.01 mole NaOH = 0.1M (use the formula above) = 10⁻¹M
[OH] = Acidity x Molarity =1 x 10⁻¹ = 10⁻¹
⇒ pOH = log₁₀[OH] = log₁₀[10⁻¹]

pH = 14 - pOH ⇒ pH = 14 - 1 = 13 (basic)