Changes into new substances called product.
Answer:
9.36
Explanation:
Sodium formate is the conjugate base of formic acid.
Also,
for sodium formate is 
Given that:
of formic acid = 
And, 
So,
Concentration = 0.35 M
HCOONa ⇒ Na⁺ + HCOO⁻
Consider the ICE take for the formate ion as:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻
At t=0 0.35 - -
At t =equilibrium (0.35-x) x x
The expression for dissociation constant of sodium formate is:
![K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5BOH%5E-%5D%5BHCOOH%5D%7D%7B%5BHCOO%5E-%5D%7D)
Solving for x, we get:
x = 0.44×10⁻⁵ M
pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64
pH + pOH = 14
So,
<u>pH = 14 - 4.64 = 9.36</u>
Answer:
19.4 g of alum, will be its theoretical yield
Explanation:
The reaction is:
2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O
Let's determine the amount of acid.
M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution
M = mmol /mL
M . mL = mmol
We replace: 8.3 mL . 9.9 M = 82.17 mmoles
We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles
Ratio is 4:2
4 moles of sulfuric acid can make 2 moles of alum
By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.
We convert moles to mass:
Molar mass of alum is: 473.52 g/mol.
0.041085 moles . 473.52 g/mol = 19.4 g
