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:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
pH: 11.8750612634
pOH: 2.12493873661
[H+]: 1.33333333333E-12
[OH-]: 0.0075
BASE
