Question

1. write the balance chemical equation for the reaction of KHP with NaOH
2. Suppose your laboratory instructor inadvertently gave you a sample of KHP contaminated with NaCL to use in standardizing your NaOH. How would this affect the molarity you calculated for your NaOH solution? Justify your answer.
3.How many grams of NaOH are needed to prepare 500ml of 0.125 M NaOH?
4 A solution of malonic acid H2C3H2O4, was standardized by titration with 0.1000 M NaoH solution. If 20.76ml of the NaOH solution is required to neutralize completely 12.95ml of the malonic acid solution, what is the molarity of the malonic acid solution.
H2C3H2O4 +2NaOH > Na2 C3H2O4 + 2H2O
5. Sodium Carbonate is a reagent that may used to standardize acids in the same way you used KHP in this experiment. In such a standardization, it was found that a 0.512g sample of sodium carbonate requried 26.30ml of sulfuric acid solution to reach the end point for the reaction.
Na2CO3(aq) + H2SO4(aq)> H2O(l) + CO2 (g) + Na2SO4(aq)
What is the molarity of the H2SO4.
6. A solution contains 6.30 ×10-2g of oxalic acid H2 C2 O4. 2H2O, in 250ml. What is the molarity of this solution.

Answer 1

1.     KHP + NaOH ---------> NaKP + H2O
2.  
3. no of moles of NaOH = molarity * volume in L
                         = 0.125*0.5 = 0.0625 moles
    mass of NaOH         = no of moles * gram molar mass
                         = 0.0625*40 = 2.5g
4.   H2C3H2O4 +2NaOH > Na2 C3H2O4 + 2H2O
     1 mole     2 mole
    H2C3H2O4                    NaOH
    M1 =                       M2 = 0.1M
    V1 = 12.95ml                V2 = 20.76ml
     n1 =1                     n2 = 2
          M1V1/n1   =    M2V2/n2
           M1       =    M2V2n1/V1n2
                    =    0.1*20.7*1/12.95*2   = 0.079M
5.Na2CO3(aq) + H2SO4(aq)> H2O(l) + CO2 (g) + Na2SO4(aq)
no of moles of Na2Co3 = W/G.M.Wt
                         = 0.512/106 = 0.0048moles
1 mole of na2Co3 react with 1 mole of H2So4
0.0048 moles of Na2Co3 react with 0.0048 moles of H2So4
   no of moles of H2So4 = molarity * volume in L
              0.0048     = molarity *0.0263
       molarity           = 0.0048/0.0263 = 0.183M
6. molarity = W*1000/G.M.Wt * volume in L
             = 0.063*1000/126*250   = 0.002M