Answer:
a. pH=1.59
Hydrofluoric acid is a weak acid, meaning it would not entirely dissociate. The capacity of dissociation of the compound will be given by it's acid dissociation constant ([tex]k_{a}[/tex]).
We know that the pH is [tex]pH=-log[H^{+} ][/tex], therefore using the dissociation constant, we can find the concentration of H+.
HF⇄ [tex]F^{-} +H^{+}[/tex]
[tex]k_{a}=\frac{[F^{-}][H^{+}]}{[HF]}[/tex]
For this acid [tex]k_{a}=6.6*10^{-4}[/tex]
Let's assume x is the unknown amount of the products, and the initial concentration, in this case 1.00M -x is the final amount of the reagent.
Therefore, [tex]k_{a}=\frac{x^{2} }{1-x}[/tex]
Neglecting the x in the numerator, given that the effect over the 1.00 M will be little we obtain that [tex]x=[H^{+}]= \sqrt{6.6*10^{-4} }=0.02569[/tex]
[tex]pH=-log(0.02569)=1.59[/tex]
b. grams NaF=34.65 g
Given the desired pH, and the initial concentration of {HF]=1.25L*1.00M=1.25 mol, we can use
[tex]pH=pk_{a}+log(\frac{[F^{-}]}{[HF]})[/tex]
[tex]3.0=-log(6.6*10^{-4})+log(\frac{[F^{-}]}{1.25})[/tex]
[tex]1.25*10^({3.0+log(6.6*10^{-4}))}=[F^{-}]=0.825 mol[/tex]
[tex]grams_{NaF}=0.825 mol F^{-}*\frac{1 mol NaF}{1 mol F^{-} }*\frac{42 g}{1mol NaF} =34.65 g[/tex]
The pH of the hydrofluoric acid solution prior to adding sodium fluoride - = 1.606 and the amount of sodium fluoride should be added to prepare the buffer solution - 35.75g or 36 g.
a) Weak acid such as hydrofluoric acid is not entirely dissociate. The capacity of dissociation of the compound will be given by it's acid dissociation constant ka
HF <==> H+ + F-
I 1.0 0 0
C -x +x +x
E 1.0-x x x
I = initial conc. C = Change in conc. E = Conc. at equilibrium
Ka = (Ka for HF is 6.3 x 10-4 constant)
= 6.3 x 10-4 =
= 6.3 x 10-4 (1.0- x ) =
Solve for x using quadratic equation method. Neglecting negative value as concentration is always positive value
x = -0.0254 and x = 0.0248 = [H+]
pH = - log [[tex]H^+[/tex]]
pH = - log 0.0248
= 1.606 ( pH of HF before adding NaF)
b) Given:
pH = 3.00
[HF] = 1.00 M
[F-] - ?
Solution:
a buffer follows the Henderson-Hasselbach equation:
pH = pKa + log
3.00 = 3.167 + log
log = -0.167
[] = (0.681) ([HF])
= (0.681) (1.00)
= 0.681 M
NaF --> + [NaF]
= 0.681 M Need grams of NaF
The amount of would be =
= [tex](41.99\ g\ NaF/ mol\ NaF)\frac{0.681\ moles\ NaF}{1 L} \times (1.25L)\ times \frac{41.99\ g\ NaF}{mol\ NaF}[/tex]
= 35.75g NaF
Thus, the pH of the hydrofluoric acid solution prior to adding sodium fluoride - = 1.606 and the amount of sodium fluoride should be added to prepare the buffer solution - 35.75g or 36 g.
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