<h3><u>Answer;</u></h3>
A) HNO3 and NO3^-
<h3><u>Explanation;</u></h3>
- <em><u>HNO3 is a strong acid and NO3 is its conjugate base, meaning it will not have any tendency to withdraw H+ from solution.</u></em>
- Buffers are often prepared by mixing a weak acid or base with a salt of that weak acid or base.
- The buffers resist changes in pH since they contain acids to neutralize OH- and a base to neutralize H+. Acid and base can not consume each other in neutralization reaction.
2NaCN(s) + H₂SO₄(aq) --> Na₂SO₄(aq) + 2HCN(g)
The molar ratio between NaCN : HCN is 2:2 or 1:1
Mass of HCN = 16.7 g
Molar mass of HCN = 1 + 12 + 14 = 27 g/mol
Molar mass of NaCN = 49 g/mol
Therefore, the mass of NaCN is
16.7 g of HCN x 49 g/mol of NaCN / 27 g/mol of HCN = 30.3 grams of NaCN
Therefore, 30.3 grams of NaCN gives the lethal dose in the room.
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>
