Question

A solution is made by mixing 13.0 g of Sr(OH)2 and 50.0 mL of 0.150 M HNO3.

A.Write a balanced equation for the reaction that occurs between the solutes.

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

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

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

E.Is the resultant solution acidic or basic?

Answer 1

A) 2 HNO3(aq) + Sr(OH)2(aq) ------------------> 2 H2O(l) + Sr(NO3)2(aq)

Step 1: Find the molar mass of HNO3:

HNO3 = 1.008 + 14.01 + 3(16) = 63.02g

Step 2: Find the mole ratio between HNO3 and Sr(NO3)2.

There is 1 mole of Sr(NO3)2 and 2 moles of HNO3, so the mole ration is 1 to 2.

Step 3: Calculate the number of Sr(NO3)2 produced.

321g HNO3 (1 mole HNO3 / 63.02g HNO3) ( 1 mol Sr(NO3)2 / 2 mole HNO3) = 2.54 moles Sr(NO3)2

First convert your 321 grams of nitric acid to moles by dividing by 63 grams per mole HNO3

Moles HNO3 = 321 over 63g/mole = 5.1 moles HNO3

Now over your 2HNO3 in the balanced equation above place 5.1 moles and under the 2HNO3 place its coefficient 2 moles

Over the Sr(NO3)2 in the same equation above place X moles and below place its understood coefficient of one mole.

Now cross multiply and solve for X moles fo strontium nitrate.

B)C)D) its resultanta mixture is a strontium nitrate its a salt so the resulthant mixture having same concentrations of oh-, sr2+ and no3-

E)   Its a salt it means a neutral comount so the resultant mixture is a neutral that equal to pH-7.