1) Given that, Ka = 2.2*10^-6 and C = 0.25 M From Ka and C, concentration of H+

ID: 699633 • Letter: 1

Question

1) Given that, Ka = 2.2*10^-6 and C = 0.25 M

From Ka and C, concentration of H+ ions in the solution can be calculated as

[H+] = sqrt(Ka*C) = sqrt(2.2*10^-6*0.25) = 0.000714 M

Now pH = -log[H+] = -log( 0.000714) = 3.12

2) The base M(OH)2 can be dissociated as

M(OH)2(s) ======= M^2+(aq) + 2OH-(aq)

Given that, pH = 10.38

From pH concentration of OH- ions produced in solution can be found as

pOH = 14 - pH = 14-10.38 = 3.62

[OH-] = 10^-pOH = 10^-3.62 = 0.00024 M

Since ionization of base M(OH)2 produces one mole of M2+ and 2 moles of OH-

If [OH-] = 0.00024 M, the concentration of M2+ = 0.00024/2 = 0.000012 M

Now, Ksp = [M2+][OH-]^2 = (0.00012)(0.00024)^2 = 6.912*10^-12

please do both parts

learning If the Ka of a monoprotic weak acid is 2.2 x106, what is the pH of a 0.25 M solution of this acid? Number pH =

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