Answer:
9.36
Explanation:
Sodium formate is the conjugate base of formic acid.
Also,
![K_a\times K_b=K_w](https://tex.z-dn.net/?f=K_a%5Ctimes%20K_b%3DK_w)
for sodium formate is ![K_b=\frac {K_w}{K_a}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%20%7BK_w%7D%7BK_a%7D)
Given that:
of formic acid = ![1.8\times 10^{-4}](https://tex.z-dn.net/?f=1.8%5Ctimes%2010%5E%7B-4%7D)
And, ![K_w=10^{-14}](https://tex.z-dn.net/?f=K_w%3D10%5E%7B-14%7D)
So,
![K_b=\frac {10^{-14}}{1.8\times 10^{-4}}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%20%7B10%5E%7B-14%7D%7D%7B1.8%5Ctimes%2010%5E%7B-4%7D%7D)
![K_b=5.5556\times 10^{-11}](https://tex.z-dn.net/?f=K_b%3D5.5556%5Ctimes%2010%5E%7B-11%7D)
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)
![5.5556\times 10^{-11}=\frac {x^2}{0.35-x}](https://tex.z-dn.net/?f=5.5556%5Ctimes%2010%5E%7B-11%7D%3D%5Cfrac%20%7Bx%5E2%7D%7B0.35-x%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>
It should be B. All of the other choices are used often by abusers
Answer:
v = 534.5mL
m = 597.15g
Density = 9.23g/mL
Density = 9.125g/mL
Explanation:
Density = mass/ volume
For the first question
Density = 1.59g/mL
Mass = 834.01g
Volume = ?
Using the above formula we have 1.59 = 834.01/v
v = 834.01/1.59
v = 534.5mL
For the second question
Density =0.9167g/mL
Volume = 651.41mL
Mass =?
Using the above formula we have
0.9167 =m/651.41
Cross multiply
m = 0.9167 x 651.41
m = 597.15g
For the third question
Mass =803.44g
Volume=87.03mL
Density =?
Density = 803.44/87.03
= 9.23g/mL
For the fourth
Density = 56.85/6.23
= 9.125g/mL
Wym kingdom? I dont get it