Question

A solution is made by mixing 12.5 g of Sr(OH)2and 40.0 mL of 0.200 M HNO3.

a. Calculate the concentration of OH− ion remaining in solution.

b.Calculate the concentration of Sr2+ ion remaining in solution.

c.Calculate the concentration of NO−3 ion remaining in solution.

d. Is the resultant solution acidic or basic?

Answer 1

Solution :-

Lets first calculate the moles of the Sr(OH)2 and HNO3

Moles = mass /molar mass

Moles of Br(OH)2 = 12.5 g /121.6347 g per mol = 0.1028 mol

Moles of HNO3 = molarity * volume in liter

                            = 0.200 mol per L * 0.040 L

                          = 0.008 mol HNO3

Balanced reaction equation

Sr(OH)2 + 2HNO3 ----- > Sr(NO3)2 + 2H2O

Moles of HNO3 are less than moles of Sr(OH)2 therefore HNO3 is the limiting reactant

Now lets calculate the moles of the Sr(OH)2 react with HNO3

0.008 mol HNO3 * 1 mol Sr(OH)2/2mol HNO3 = 0.004 mol HNO3

Therefore moles of Sr(OH)2 reacted = 0.1028 mol – 0.004 mol = 0.0988 mol

So moles of OH- remain = 0.0988 mol *2 = 0.1976 mol OH-

a)Therefore concentration of OH- = 0.1976 mol / 0.040 L = 4.94 M OH-

b) Concentration of the Sr^2+ = 0.1028 mol / 0.040 L =    2.57 M Sr^2+

c) Concentration of the NO3^- = 0.008 mol / 0.040 L = 0.200 M

d) Since the OH- concentration is higher therefore resulting solution is basic in nature