1. A sample of potassium hydrogen iodate (KH(IO 3 ) 2 ), which is an acid, with

Question

1. A sample of potassium hydrogen iodate (KH(IO3)2), which is an acid, with a mass of 0.5608 g is dissolved in water and used to standardize a solution of potassium hydroxide. An endpoint is reached when 45.47 mL of potassium hydroxide has been added. Determine the molarity (in mol/L) of the potassium hydroxide solution. Report your answer to four significant figures.

2. The standardized potassium hydroxide solution is then used to titrate 36.53 mL of a solution of sodium hydrogen citrate (Na2C3H5O(COOH)(COO)2). An endpoint is reached when 27.45 mL of potassium hydroxide has been added. Determine the molarity (in mol/L) of the sodium hydrogen citrate solution. Report your answer to four significant figures.

Explanation / Answer

1.

   KH(IO3)2(aq) + KOH(aq) ---> K2(IO3)2(aq) + H2O(l)

1 mol KH(IO3)2(aq) = 1 mol KOH(aq)

so that,

nO of mol of KH(IO3)2 taken = weight/Molecular weight

                = 0.5608/389.92

                = 0.00144 mol

No of mol of KOH reacted = 0.00144 mol

Molarity of KOH reacted = n/V

                        = 0.00144/(45.47*10^-3)

           = 0.03167 M

2.

Na2C3H5O(COOH)(COO)2(aq) + KOH(aq) ---> Na2C3H5O(COOK)(COO)2(aq) + H2O(l)

1 mol Na2C3H5O(COOH)(COO)2(aq) = 1 mol KOH(aq)

No of mol of KOH reacted = M*V

           = 0.03167*27.45*10^-3

           = 0.00087 mol

No of mol of Na2C3H5O(COOH)(COO)2 reacted = 0.00087 mol

Molarity of Na2C3H5O(COOH)(COO)2 reacted = n/v

                   = 0.00087/(36.53*10^-3)
              
                   = 0.02382 M

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