A 5.00-mL sample of nitric acid required 13.25 mL of 0.0842 M strontium hydroxid

Question

A 5.00-mL sample of nitric acid required 13.25 mL of 0.0842 M strontium hydroxide for titration. Calculate e molarity of the acid solution. A solution made by adding 9.00 g of phosphoric acid to 90.00 mL of water had a volume of 93.60 mL. Calculate the molarity of this solution. A solution was made by adding water to 0.250 g of H_2 Z until the volume totaled 25.00 mL. Subsequent titration required 40.50 mL of 0.100 M sodium hydroxide. Calculate the molar mass of H_2 Z (in g/mol). given: H_2 Z(aq) + 2 NaOH(aq) rightarrow Na_2 Z(aq) + 2 H)2O(l)

Explanation / Answer

1. sr(OH)2 + 2HNO3 ---> Sr(NO3)2 + 2H2O

from equation ;

1 mol sr(OH)2 = 2 mol HNO3

No of mol of sr(OH)2 = 0.01325*0.0842 = 0.00111565 mol

No of mol of HNO3 = 0.00111565*2 = 0.0022313 mol

concentration of HNO3 = 0.0022313 /0.005 = 0.44626 M

2. No of mol ofH3PO4 = 9/98 = 0.092 mol

Molarity = n/V in L = 0.092/0.09 = 1.02 M

3. No of mol of NaOH = 0.04*0.1 = 0.004 mol

   No of mol of H2Z = 0.004/2 = 0.002 mol

molarmass of H2Z = 0.25/0.002 = 125 g/mol

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