An experiment calls for you to use 200. mL 1.0 M HNO_3 solution. All you have av

Question

An experiment calls for you to use 200. mL 1.0 M HNO_3 solution. All you have available is a bottle of 6.0 M HNO_3 How would you prepare the desired solution? (MM: HNO_3 = 63.02 g/mol) In the laboratory 6.67 g of Sr(NO_3)_2 in dissolved in enough water to form 0.750 L. A 0.100 L sample is withdrawn from this stock solution and titrated with a 0.0460 M solution of Na_3PO_4. a. What is the concentration of the Sr(NO_3)_2 stock solution? b. Write a balanced molecular equation for the titration reaction. C. How many millilitiers of the Na_3PO_4 Solution are required to all the Sr^2+ ions in the 0.100 L sample? (MM's: Sr(NO_3)_2 = 211.64; Na_3PO_4 = 163.94)

Explanation / Answer

Ans. 1.

M1V1 = M2V2                       - equation 1

M1= molarity of solution 1, V1= volume of solution 1          ; [Stock HNO3, 6.0 M]

M2= molarity of solution 2, V2= volume of solution 2         ; [Final solution, 1.0 M]

Given,

            M1 = 6.0 M                 ; V1= ?

            M2 = 1.0 M                 ; V2 = 200.0 mL

Putting the values in equation 1-

            6.0 M x V1 = 1.0 M x 200.0 mL

            Or, V1 = (1.0 M x 200.0 mL) / 6.0 M = 33.33 mL

Preparation: Take 33.33 mL of stock HNO3 in a 200.0- mL standard volumetric flask. Make the final volume upto the mark with distilled water. The resultant solution is 1.0 M, 200.0 mL.

Ans. 2.a. Moles of Sr(NO3)2 = Mass/ molar mass

                                                = 6.67 g / (211.64 g/mol)

                                                = 0.0315 mol

Concertation of Sr(NO3)2 solution = number of moles / volume of solution in L

                                                = 0.0315 mol / 0.750 L

                                                = 0.0420 mol/ L

                                                = 0.0420 M

Thus, Concertation of resultant Sr(NO3)2 stock solution = 0.0420 M

2A. Balanced reaction-        2 Na3PO4(aq) + 3 Sr(NO3)2(aq) = 6 NaNO3(aq) + Sr3(PO4)2(s)

Stoichiometry: 2 mol Na3PO4 reacts with 3 mol Sr(NO3)2 to form 6 mol NaNO3 and 1 mol Sr3(PO4)2.            

2C. Sr2+ is precipitated as Sr3(PO4)2(s)

Stoichiometry: 2 mol Na3PO4 is required per 3 mol Sr(NO3)2 for complete precipitation.

1 mol Sr(NO3)2 has 1 mol Sr2+.

So,

Moles of Sr2+ in sample = Molarity of solution x Volume (in L) taken for titration

                                    = 0.0420 M x 0.100 L

                                    = 0.00420 moles

Now, using stoichiometry :

            3 mol Sr(NO3)2 is precipitated by 2 mol Na3PO4

     Or, 1 mol   -           -           -           -           (2/3) mol Na3PO4

     Or. 0.0042 mol -        -        -        (2/3) x 0.0042 mol Na3PO4

                                                                        = 0.0280 mol Na3PO4

Thus, number of moles of Na3PO4 required for complete precipitation = 0.0280 mol

Required volume (in L) of Na3PO4 =number of moles required / Molarity of solution

                                                = 0.0280 mol / 0.0460 M

                                                = 0.60900 L

                                                = 609.00 mL                                      ; [1 L = 1000 mL]

Thus, required volume of 0.0460 M Na3PO4 solution = 609.00 mL    

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