Calculate the ph of a 1.5 M weak acid solution. Ka = 3.0 X 10-6?

Question

solve this if you are good at chemistry!

Answer 1

Actually we can't solve it if you don't provide the valency of the acid... Is it monoprotic, diprotic, triprotic or what?

For a monoprotic acid let's assume the formula HA, then

.. .. .. .. .. .. ..HA <=> H+ + A-
Initial .. . .. .. 1.5
Dissociate .. x
Produce .. .. .. .. .. .. ..x .. .. x
At Equil .. .. 1.5-x .. .. ..x .. ..x

Ka= [H+][A-]/[HA] = x^2/(1.5-x)
The best thing is to solve the quadratic.
However for simplicity we assume that 1.5 >> x so practically 1.5-x=1.5
thus
Ka=x^2/1.5 => x=squareroot(Ka*1.5) =SQRT(1.5*3*10^-6) = 2.1*10^-3 which is <<1.5 so our assumption is valid (otherwise you would have to solve the quadratic)

pH=-log[H+] =-logx = -log(2.1*10^-3) =2.68

Answer 2

(a million) enable HA = acetic acid and A- = acetate ion A- + H2O --> HA + OH- Ka for acetic acid = a million.75x10^-5 Kb = [HA][OH-] / [A-] Kb = Kw/Ka = 1x10^-14 / a million.75x10^-5 Kb = 5.714x10^-10 enable X = [OH-] = [HA] 5.714x10^-10 = X^2 / (0.38 - X) X<<0.38 X = a million.4735x10^-5 = [OH-] pOH = 4.80 3 pH = 14 - pOH = 9.17 (4) enable HA = benzoic acid Ka = 6.3x10^-5 = [H+][A-] / [HA] enable X = [H+] = [A-] Ka = 6.3x10^-5 = X^2 / (0.00014 - X) in view that X is only no longer << 0.00014, you will would desire to remedy the quadratic equation. remedy for X and divide by employing 0.00014 to get the fraction ionized. (3) this is similar to (a million) B + H2O --> OH^-a million + BH^+a million you p.c. Kb for pyridine enable X = [OH-] = [BH+] Kb = X^2 / 0.650 X<<0.650 so Kb = X^2 / 0.650 X = [OH] and -log[OH] = pOH pH = 14 - pOH

Answer 3

Using Ostward Dilution Law , assuming the degree of dissociation is small , then

[H+] = Sqrt[ Ka . c ] where c = concentration of the acid solution .

Then use pH = - log[H+]