I have no idea how to do these problems

Question

1. In a titration,, 50.0 mL of strontium hydroxide, Sr(OH)2, is neutralized with 34.8 mL of 0.20 M 

sulfuric acid solution? What is the concentration of strontium hydroxide

2.  In a titration of 50.00 mL of 0.200 M HNO3, what volume of a 0.100 M KOH solution is required for 

     neutralization?

(A) 50.00 mL

(B) 25.00 mL

(C) 75.00 mL

(D) 100.0 mL

Answer 1

1.) Write the equation for the titration:

Sr(OH)₂ + H₂SO₄ ==> SrSO₄ + 2H₂O

To find [Sr(OH)₂], we need to find moles of Sr(OH)₂ and divide that by 0.05 L (50.0 ml)

moles H₂SO₄ = 34.8 ml x 1 L/1000 ml x 0.20 mol/L = 0.00696 moles H₂SO₄

moles Sr(OH)₂ = 0.00696 moles because in the equation there is a 1:1 mol ratio of H₂SO₄ : Sr(OH)₂

[Sr(OH)₂] = 0.00696 moles/0.05 L = 0.1392 M = 0.140 M (3 sig. figs.)

2.) Write the equation for the titration:

HNO₃ + KOH ==> KNO₃ + H₂O

moles HNO₃ = 0.050 L x 0.200 mol/L = 0.010 moles

moles KOH needed = 0.010 moles KOH because they react in a 1:1 mole ratio

volume KOH needed: (x L)(0.100 mol/L) = 0.01 moles

x = 0.1 L = 100. ml

Answer = D (except there are too many sig figs in 100.0 ml so even that answer is incorrect. There should be only 3 sig. figs as in 100. ml)

Answer 2

1. We know, S₁V₁= S₂V₂

here, S₁= Concentration of Sr(OH)₂

V₁= Volume of Sr(OH)₂

S₂= Concentration of sulfuric acid

V₂= Volume of sulfuric acid

So, S₁ x 50.0 mL = 0.20 M x 34.8 mL

S₁= (0.20 M x 34.8 mL)/50.0 mL = 0.1392 M

2) We know, S₁V₁= S₂V₂

here, S₁= Concentration of KOH = 0.100 M

V₁= Volume of KOH

S₂= Concentration of HNO₃ = 0.200 M

V₂= Volume of HNO₃ = 50.00 mL

So, 0.100 M x V₁ = 0.200 M x 50.00 mL

V₁ = ( 0.200 M x 50.00 mL)/ 0.100 M = 100 mL