50.0L of 0.185M Sr(OH)2 is reacted with 35.0mL of a solution containing 0.130g of HCl.

Question

a. Calculate the pOH resulting from the mixture

b. suggest an indicator that will be effective for this incomplete titration and explain your choice

I'm just having trouble setting everything up.

Answer 1

You should ALWAYS write down the balanced equation:

Sr(OH)₂ + 2HCl ==> SrCl₂ + 2H₂O

Next, we want to calculate the moles of each reactant so that we can determine which one is limiting:

moles Sr(OH)₂ = 50.0 ml x 1 L/1000 ml x 0.185 mol/L = 0.00925 mol Sr(OH)₂

moles HCl = 0.130 g HCl x 1 mol HCl/36.5 g = 0.00356 mol HCl

Since it takes 2 mol HCl per mol Sr(OH)₂, HCl is clearly in limited supply and is limiting.

This means Sr(OH)2 will be left over after the reaction is complete. Also note that the final volume will be

50.0 ml + 35.0 ml = 85.0 ml = 0.0850 L

Now, we can calculate how much Sr(OH)₂ is left over:

0.00356 mol HCl x 1 mol Sr(OH)₂/2 mol HCl = 0.00178 moles Sr(OH)₂ used up

0.00925 moles - 0.00178 moles = 0.00747 moles Sr(OH)₂ remaining

Recall that this is now in a volume of 0.0850 L

[Sr(OH)₂] = 0.00747 mol/0.0850 L = 0.0879 M

Since [OH-] is twice that of Sr(OH)₂ b/c it dissociates ==> Sr²⁺ + 2OH-

[OH-] = 2 x 0.0879 = 0.176 M

pOH = - log [OH-] = -log 0.176

pOH = 0.754

To choose a suitable indicator, you want one that has a pKa close to the pH at the end point. If you have a list, choose from that list. If not, a good one might be Indigo Carmine