If the Na2CO3 weight taken for the titration was 0.4444 gram and the concentrati

Question

If the Na2CO3 weight taken for the titration was 0.4444 gram and the concentration of the HCl solution was 0.096M, calculate the equivalence point volume expected. SHOW ALL WORK CLEARLY. (Formula weights: Na2CO3 = 105.99 gram/mole; HCl = 36.46 gram/mole)

Explanation / Answer

moles of Na2CO3 = 0.444 g/ 105.99 g/mole = 0.00419 mol. At equivalent point moles of HCl used will be equal to moles of Na2CO3 but we need two moles of HCl per mole of Na2CO3 therefore moles of HCl is 2 * 0.00419 mol = 0.00838 mol. Now calculate the volume of 0.096M HCl solution containing 0.00838 mol. -----> 1000 mL of HCl = 0.096 mol thus 1 mole of HCl is present in 1000/ 0.096 mL. Therefore 0.00838 mol of HCl will be present in 0.00838 * 1000/ 0.096 mL = 87.29 mL. The above post at 12 min is also mine I just forgot to multiply the mole number of HCl by two. this is the correct answer.

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