Question

1.Calculate the pH of a solution of 4.7M H2SO4 (Diprotic but strong acid). Look up Ka/pKa online.

2.What is the pH and pOH of 0.24M solution of Ethanolamine. Look up Ka/pKa.

Please show all work. I want to learn how to do it not just have the answer. Thank you!

Answer 1

H2SO4 is a storng acid as its pKa1 value is -3 or Ka1 = 10^-pKa1 = 10³ and pKa2 = 1.99 or Ka2 = 10^-pKa2 = 10^-1.99

So, compelete dissociation of H2SO4 -

H2SO4 <-> 2H+ + SO42-

I 4.7 0 0

C -4.7x +9.4x +9.4x

E 4.7-4.7x 9.4x 9.4x

K = Ka1*Ka2 = 10³*10^-1.99 = 10^1.01 = 10.2329

Also K = [H+]²[SO42-]/[H2SO4] = (9.4x)²/(4.7-4.7x) = 10.2329

so, K = 18.8x²/(1-x) = 10.2329

or 18.8x² = 10.2329 - 10.2329x

so, 18.8x² + 10.2329x - 10.2329 = 0

so, x = (-10.2329 + sqrt(10.2329² - 4*18.8*(-10.2329))/(2*18.8)

so, x = 0.514

so, H+ = 9.4x = 9.4*0.514 = 4.8316 M

so pH = -log(H+) = -log(4.8316) = -0.684

2.

Ethanolamine is a weak base as its pKa value is 9.5 so, its pKb value is 14-9.5 = 4.5 or Kb = 10^-pKb = 10^-4.5

Ethanolamine has formula H2N- CH2-CH2OH

When it attracts an H+ ion from H2O it forms an equilibrium like-

H2N- CH2-CH2OH + H2O -> (H3N+)- CH2-CH2OH + OH-

So, Kb = [(H3N+)- CH2-CH2OH][OH-]/[H2O][H2N- CH2-CH2OH]

Making ICE table -

H2N- CH2-CH2OH + H2O -> (H3N+)- CH2-CH2OH + OH-

I 0.24 0 0

C -0.24x +0.24x +0.24x

E 0.24-0.24x 0.24x 0.24x

So, Kb = (0.24x)²/(0.24-0.24x) = 10^-4.5

So, 0.24x²/(1-x) = 10^-4.5 = 3.16*10^-5

Assuming x<<1 and thus neglecting it compared to 1 we get

0.24x² = 3.16*10^-5

so, x = sqrt(3.16*10^-5/0.24) = sqrt(13.17*10^-5) = 1.1476 * 10^-2

so, OH- = 0.24x = 0.24*1.1476*10^-2 =  0.275*10^-2

So, pOH = -log(OH-) = -log(0.275*10^-2) = 2.56

so, pH = 14-pOH = 14 - 2.56 = 11.44