Question

A 65.1 ml solution of perchloric acid has a ph of 3.461. determine the initial molarity of the strontium hydroxide slution that was added to the 65.10 ml perchloric acid to obtain a final 100 ml solution with a ph of 6.37.

Answer 1

perchloric acid is a very strong acid, it has the formula of HClO4(aq)

when added with strontium hydroxide, Sr(OH)2 we will have the next equation/reaction

HClO4(aq) + Sr(OH)2(aq) = H2O(l) + SrClO4(aq)

balance:

2HClO4(aq) + Sr(OH)2(aq) = 2H2O(l) + Sr(ClO4)2(aq)

note that we have a 1:2 ratio

so...

initially:

relate HClO4 concentration to H+

[H+] = 10^-pH

[H+] = 10^-3.461 = 0.00034593

so...

[HClO4] = [H+] = 0.00034593

calculate mol of H+

mol of H+ = MV = 0.00034593*65.1/1000 = 0.00002252004

after adding Sr(OH)2

we get

[H+] = 10^-pH = 10^-6.37 =4.265*10^-7

V = 100 mL = 0.1 L

mol of H+ = MVtotal = (4.265*10^-7)(0.100) = 4.265*10^-8 M of H+

change in moles of H+:

dH+ = (0.00002252004-4.265*10^-8) = 0.00002247739

so...

mmol of OH- --> 0.00002247739

relate to 1:2 to Sr(OH)2 : OH-

0.00002247739/2 = 0.00001123869 mol of Sr(OH)2

now...

initial molarity of Sr(OH)2 when added to 65.10 mL

Vbase = Vtotal - Vacid = 100-65.10 = 34.9 mL = 34.9/1000 = 0.0349 L

[Sr(OH)2] = mol of  Sr(OH)2 / total V = 0.00001123869 / (0.0349) = 0.000322 M of Sr(OH)2