Calculate the ph of a 3.78×10-3 m solution of naf, given that the ka of hf = 6.80 x 10-4 at 25°c.

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30-03-2018
Chemistry

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Calculate the ph of a 3.78×10-3 m solution of naf, given that the ka of hf = 6.80 x 10-4 at 25°c.

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zoexoe zoexoe · Answer 1 · 2018-04-08T11:35:07+02:00

The answer is pH = 7.37.

Solution:
First, we set up an ICE table for the reaction:
F- + H2O → HF + OH-
Initial 0.00378 0 0
Change -x +x +x
Equilibrium 0.00378-x x x

We can calculate Kb from the given Ka, since we know that Kw = Ka*Kb = 1.0 x 10^-14 at 25°C:
Kb = Kw/Ka = 1.0x10^-14 / 6.80x10^-4 = 1.471 x 10^-11
Kb = 1.471 x 10^-11 = [OH-][HF] / [F-] = (x)(x) / (0.00378-x)

Approximating that x is negligible compared to 0.00378 simplifies the equation to
1.471x10^-11 = (x)(x) / 0.00378
1.471 x 10-11 = x2 / 0.00378

Then we solve for x that is also equal to [OH-]:
x2 = (1.471 x 10^-11)(0.00378)
x = sqrt[(1.471 x 10^-11)(0.00378)] = 2.358x10^-7 = [OH-]
in which 0.0000002358 is indeed negligible compared to 0.00378.

We can now calculate for pOH:
pOH = -log [OH-] = -log (2.358x10^-7) = 6.63

Therefore, the pH is
pH = 14 - pOH = 14 - 6.63 = 7.37

mahima6824soni mahima6824soni · Answer 2 · 2022-03-09T08:17:04+01:00

The pH of a 3.78 × 10-3 m solution of naf is 7.37.

What is pH?

PH is the measurement of acidic and basic of any liquid. pH means potential hydrogen.

First, we calculate an ICE table for the reaction

[tex]F^- + H_2O = HF + OH^-[/tex]

Initial 0.00378 0 0

Change -x +x +x

Equilibrium 0.00378-x x x

Now, calculate the Kb from the Ka.

We know that,

[tex]Kw = Ka \times Kb = 1.0 \times 10^-14\;at\; 25^\circ C[/tex]

[tex]Kb = \dfrac{Kw}{Ka } = \dfrac{1.0\times 10^-14}{6.80\times10^-4} = 1.471 \times 10^-11\\\\\\ Kb = 1.471 \times 10^-11 =\dfrac{ [OH-][HF]}{ [F-]} = \dfrac{ (x)(x)}{(0.00378-x)}[/tex]

Assume that x is negligible compared to 0.00378

[tex]1.471 \times10^-11 = \dfrac{(x)(x)}{0.00378} \\\\ 1.471 \times 10-11 = \dfrac{x2}{0.00378 }[/tex]

Now, calculate the x for OH

[tex]x2 = (1.471 \times 10^-11) (0.00378) \\\\ x = [(1.471 x 10^-11)(0.00378)]^2 = 2.358\times 10^-7 = [OH-][/tex]

Calculating the pOH

[tex]pOH = -log [OH-] = -log (2.358\times 10^-7) = 6.63[/tex]

pH = 14 - pOH =

pH = 14 - 6,63 = 7.37

Thus, the pH is 7.37.

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