Question

4. Test your understanding of titrations. a) Write a balanced equation for the neutralization of aqueous strontium hydroxide with aqueous hydrochloric acid. b) A chemist finds a strontium hydroxide solution in lab, but the concentration of this solution is not labeled. A 25.00 ml sample of strontium hydroxide solution is placed in a flask. It takes the addition of 30.00 mL of 0.2695 M HCl to reach the equivalence point. i. Define equivalence point. ii. What is the molarity of the strontium hydroxide solution? (0.1617 M] iii. What is the molarity of aqueous Hions in the 0.2695 M HCl solution used in this titration? (Note- Aqueous H+ ions are also known as H3O+ ions) (0.2695 M) iv. What is the pH of the aqueous HCl solution? Include correct significant figures! (pH=0.5694)

Answer 1

a)
Sr(OH)2 (aq) + 2 HCl (aq) --> SrCl2 (aq) + 2H2O (l)

b) Equivalence point is that point in the titration where
volume of acid added just enough to neutralize the base
taken.

c)At equivalence point , moles of base present = moles of acid used

moles of HCl present = volume * molarity
= 30 ml * 0.2695 M
= = 8.085 milli moles

balanced equation says for 1 mole S(OH)2 , 2 mole HCl needed

Thus moles of Sr(OH)2 present is = 1/2*8.085 milli moles
= 4.0425 milli mols
molarity of Sr(OH)2 = # of moles of Sr(OH)2/volume of Sr(OH)2
= 4.0425 milli mols/25 ml = 0.1617 M

Answer : 0.1617 M
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3)
Since HCl is a strong acid , it will ionize completely
HCl --> H+ + Cl-
so molarity of HCl = molarity of H+

[HCl] = [H+] = 0.2695 M
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4)
pH = -log[H+]
= -log(0.2695)
=0.5694
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All solved . kindly give a thumbs up :)