A is a solution of 0.01molperdm³, HCL.
B is a solution of sodium hydroxide
I. Calculate the average titre value
II. Calculate the concentration of B in mol/dm³
III. Concentration of B in grams per dm³
IV. The number of moles of acid in average titre
V. The number of moles of alkaline in the volume of B pipette .
Reaction between NaOH and HCl can be given as follows "NaOH+ HCl \\longrightarrow NaCl+ H_2O"
Molarity of HCl is 0.01 M, mole of HCl react with NaOH = 0.01
(i) Average titrate value "= \\frac{V_1+V_2}{2}= \\frac{0.01+0.01}{2}=0.01"
(ii) concentration of B = 0.01 molar
(iii) HCl is acid and its molarity is 0.01 so number of moles of acid is 0.01
(iv) Since molarity is same for both then mole of alkalines =0.01
Comments
Leave a comment