Answer:
The answer is 5.10
Explanation:
<h3><u>Given</u>;</h3>
<h3>
<u>To </u><u>Find</u>;</h3>
We know that
pH + pOH = 7
pOH = 7 – pH
pOH = 7 – 1.90
pOH = 5.10
Thus, The pOH of the solution is 5.10
A)
NH⁴⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃0⁺<span>(aq)
- acid </span>a species that able to donate (H+): NH⁴⁺
- base a species that is able to accept a proton (H+): H₂O
- conjugate base a species formed when acid donates a proton (H+): NH₃
- conjugate acid a species formed by a base accepts a proton (H+): H₃0⁺
b)
CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)
- base a species that is able to accept a proton (H+): CN⁻
- acid a species that able to donate (H+): H₂O
- conjugate acid a species formed by a base accepts a proton (H+): HCN
- conjugate base a species formed when acid donates a proton (H+): OH⁻
Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T =
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where,
= concentration = 
Hence, putting the given values into the above equation as follows.
=
=
=
m
or, =
= 1 nm (approx)
Also, it is known that
= 
Hence, we can conclude that addition of 0.1
of KCl in 0.1
of NaBr "
" will decrease but not significantly.