A 0.1195g sample of NA2CO3 was added to a 250ml flask, dissolved in 35ml of boil

Question

A 0.1195g sample of NA2CO3 was added to a 250ml flask, dissolved in 35ml of boiled distilled water and 5 drops of phenolphthaline indicator added. Next, 40.00ml of an unknown HCl solution was added to the flask, turning the solution colorless indicating excess acid was present. The resultion solution was boiled for 2-3 mins. It the required 17.51ml of 0.1096M NaOH to back titrate the excess H+ in solution to the faint pink phenophthaline endpoint. Calculate the concentration of the unknown HCl solution. The Answer is 0.1044 Can anybody show me how to get to this value?

Explanation / Answer

Na2CO3 + H2O ----> 2NaOH + CO2 (when heated) , 0.1195 gm Na2CO3 = 0.1195/105.98 = 0.0011276 moles Na2CO3, moles of NaOH produced = 0.0011276 x2 = 0.002255 , , NaOH moles later added = ( 17.51 x.1096/1000) = 0.001919 , total NaOH moles = 0.002255+0.001919 = 0.004174 , moles of HCl = 0.004174 at neutralization point , Molarity of acid = moles/vol = 0.004174 x1000/40 = 0.1044 , hence conc of acid = 0.1044 M

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