Question

A 0.070 M solution of the salt NaB has a pH of 9.60. Calculate the pH of a 0.010 M solution of HB.

Or, what is the Kb of HB?

Answer 1

Answer:

pH=3.68

Explanation:

Hello,

At first, by knowing the 9.60-pH of the 0.070M solution of NaB, we can compute the Kb as the B contained into the NaB behaves as a base:

$B^-+H_2O(l)<-->HB+OH^-$

Now, one can compute concentration of the OH ions because it is the same concentration of the HB based on the aforementioned chemical reaction:

$pH=-log([H^+])\\\\ H^+=10^{-pH}=10^{-9.60}=2.51x10^{-10}\\OH^-=\frac{Kw}{[H^+]} =\frac{1x10^{-14}}{2.51x10^{-10}} =3.98x10^{-5}$

$[OH^-]=[HB]=3.98x10^{-5}M$

The Kb is then:

$Kb=\frac{[HB][OH^-]}{[B^-]}=\frac{(3.98x10^{-5})^2}{0.070-3.98x10^{-5}} =2.264x10^{-8}$

After doing that, the Ka for the acid, taking into account its dissociation is:

$HB<-->H^++B^-\\Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{2.264x10^{-8}} =4.42x10^{-7}$

Based on the ICE conditions and table, one states the change during the dissociation of HB:

$Ka=\frac{[H^+][B^-]}{[HB]}\\4.42x10^{-6}=\frac{x^2}{0.010-x} \\x=2.08x10^-4M$

Finally, the found value for x equals to the H+ concentration, so we compute the pH:

$[H^+]=2.08x10^{-4}\\pH=-log(2.08x10^{-4})\\pH=3.68$

Best regards.

Answer 2

Th pH of a 0.010 M solution of HB is 3.68 and value of Kb for HB is 2.26×10⁻⁸.

How de we calculate pH?

pH of any solution is define as the negative logarithm of the concentration of H⁺ ion in the solution.

Given that, pH = 9.6

9.6 = -log[H⁺]

[H⁺] = $10^{-9.6}$ = 2.51×10⁻¹⁰

Also we know that,

[H⁺][OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.51×10⁻¹⁰ = 3.98×10⁻⁵

Now we calculate the value of Kb for HB following the below equation:
                                     B⁻  +  H₂O(l)  →  HB  +  OH⁻

Initial:                        0.070                      0         0

Equilibrium:        0.070-3.98×10⁻⁵   3.98×10⁻⁵ 3.98×10⁻⁵

Kb = [3.98×10⁻⁵]² / (0.070-3.98×10⁻⁵)

Kb = 2.26 × 10⁻⁸

Also we know the relation Ka × Kb = 10⁻¹⁴

Ka = 10⁻¹⁴/2.26 × 10⁻⁸ = 4.42×10⁻⁷

For the below equation, value of Ka will be:

                              HB  ↔  H⁺  +  B⁻

Initial:                   0.010      0       0

Equilibrium:        0.010-x    x        x

Ka = x² / (0.010-x)

We can neglect the value of x as it is very small as compare to 0.010.

4.42×10⁻⁷ = x² / 0.010

x = 2.08×10⁻⁴ M

So, the pH of the HB solution is:

pH = -log(2.08×10⁻⁴)

pH = 3.68

Hence required pH is 3.68.

To know more about Kb & Ka, visit the below link:

https://brainly.com/question/26998