A 25. 0-ml sample of 0. 150 m hydrocyanic acid is titrated with a 0. 150 m naoh solution. What is the ph beforeany base is added
1 answer:
You might be interested in
Explanation:
The given data is as follows.
[HCOOH] = 0.2 M, [NaOH] = 2.0 M,
V = 500 ml, [Benzoic acid] = 0.2 M
First, we will calculate the number of moles of benzoic acid as follows.
No. of moles of benzoic acid = Molarity × Volume
=
= 0.095 mol
And, moles of NaOH present in the solution will be as follows.
No. of moles of NaOH = Molarity × Volume
=
= 0.05 mol
Hence, the ICE table for the chemical equation will be as follows.
Initial: 0.095 0.05 0 0
Equlbm: (0.095 - 0.05) 0 0.05
pH =
=
= 4.245
For,
Initial: 0.2x 2(0.5 - x) 0
Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)
As,
pH =
4.245 = 3.75 +
= 0.5
= 3.162
Now,
= 3.162
x = 0.464 L
Volume of NaOH = (0.5 - 0.464) L
= 0.036 L
= 36 ml (as 1 L = 1000 mL)
And, volume of formic acid is 464 mL.
Answer:
2.52L
Explanation:
Given parameters:
T₁ = 400K
V₁ = 4L
T₂ = 252K
unknown
V₂ = ?
Solution:
To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.
Mathematically,
Substitute and solve for V₂
V₂ = 2.52L