Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:
Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
![\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}](https://tex.z-dn.net/?f=%5CKa%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BHCOO%5E-%5D%7D%7B%5BHCOOH%5D%7D)
From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,
Let's solve this for x. Multiply both sides by 0.12
taking square root to both sides:
Now, we have got the concentration of ![[H^+] .](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20.)
![[H^+] = 0.00465 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%200.00465%20M)
We know that, ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
1. Q=112.8 kJ
2. Q=5.01 kJ
<h3>Further explanation</h3>
The heat required for phase change :
Q = mLf
Lf=latent heat of fusion
- vaporization/condensation
Q = mLv
Lv=latent heat of vaporization
1.
m=50 g=0.05 kg
Lv (water) = 2256 kJ/kg
2.
m=15 g=0.015 kg
Lf for water = 334 kj/kg
Hello.
He have that:
[H+][OH-] = 10⁻¹⁴
3.64 x 10⁻⁸ [OH-] = 10⁻¹⁴
[OH-] = 0.27 x 10⁻⁶ mol/L
M₁V₁ = M₂V₂
(0.050M)*(10 mL) = (0.10M)*(x)
x = 5 mL
You would take 5 mL of the 0.10 M solution of HCl and add 5 mL of deionized water. Therefore, the total volume will be 10 mL and it will have a molarity of 0.050.
Answer:
154 kg
Explanation:
Step 1: Define conversion
1 kg = 2.20 lbs
Step 2: Use Dimensional Analysis
= 153.636 kg
Step 3: Simplify
We are give 3 sig figs.
153.636 kg ≈ 154 kg
