s. Calculate the K, for an unknown acid Hx, siven thast a 0.25 M salt solution N

Question

s. Calculate the K, for an unknown acid Hx, siven thast a 0.25 M salt solution Nax has a pHl of.s e.43 () 0.as-x 6. Calculate the K, for an unknown base, B, given that a 0.400 M solution of the salt HBCI has a percent dissociation of 0.030%. 7. Calculate the pH of a solution created by mixing 50.00 mL of 0.200 M HF with 25.00 ml of 0250 M NaF. 8. Calculate the pH of a solution created by mixing 5.50 g NH,CI into 100.00 mL of 048 M NH (ignore any volume changes). c. 1.4x 10, d. 1.0x 10,e2.4x 10 Ja 0.015 M,0010 M, 6 1 10.32, e.690, £.7.309.56, 5 27x10, 628x la

Explanation / Answer

5)

we have below equation to be used:

pH = -log [H+]

8.48 = -log [H+]

log [H+] = -8.48

[H+] = 10^(-8.48)

[H+] = 3.311*10^-9 M

we have below equation to be used:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(3.311*10^-9)

[OH-] = 3.02*10^-6 M

Lets write the dissociation equation of X-

X- +H2O -----> HX + OH-

0.25 0 0

0.25-x x x

Kb = [HX][OH-]/[X-]

Kb = x*x/(c-x)

Kb = 3.02*10^-6*3.02*10^-6/(0.25-3.02*10^-6)

Kb = 3.648*10^-11

we have below equation to be used:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/3.648*10^-11

Ka = 2.74*10^-4

Answer: 2.74*10^-4

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