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>
If we fertilize a plant, then its height increases fast. Always use if then format
Yes, o-toluic acid is soluble in ether as ether is slightly polar and it is soluble in NaOH because it is likely to form soluble compounds with it.
Naphthalene is insoluble in NaOH.
Answer:
Explanation:
according to balance chemical equation
3 A2 moles produced 2 moles of A3B
so 12 moles A2 will produced moles of A3B= 12*2/3=24/3= 8
therefore 12 moles of A2 produced 8 moles of A3B
Answer:
92.6
Explanation:
6 mol x 18.02 g of H2o --> 3 mol x 58.33 g Mg(OH)2
108.12 g of h2o --> 174.99 of Mg(OH)2
g of H2O is 150 g of Mg(OH)2
150g x 108.12g / 174.99 =
92.67
