e: Part B: Determination of the Molar Mass of a Solid Acid Date: nknown NumberAU

Question

e: Part B: Determination of the Molar Mass of a Solid Acid Date: nknown NumberAU6S Number of moles of ionizable H" ions per mole of solid acid MoH 040 93Mhe average molarity of NaOH from Part A.) Data: Trial 1 Trial 2 Mass of sample Initial volume, NaOH solution Final volume, NaOH solution 9 mL Calculations Trial 1 Trial 2 Volume of NaOH solution used 000311 mo525 mol Moles of NaOH (OH) used Moles of H in sample Moles of solid acid in sample Molar mass of solid acid mol mol mol g/mol _ gmol gmol Average Molar Mass 'm not reaily sure how to find the ansners to the reol of the questioron this labso please help. Asotve equation above does go of muy nknown Sokd Page 7 of 8 Thank o

Explanation / Answer

answer=

equation like ==>H3po4+3NaOH==>Na3po4+3H2O here acid have 3 basicity...so it can donate 3 H+ ion.

moles of H+ ion ===> always be equal to moles of NaOH because at neutral point both have same mole to neutral each other..so its value ===>trial 1=0.00317 and for trial 2=0.00328

moles of solid acid ===> its one mole contain 3 mole H+ ion so here it value ==>trial 1 =0.00317/3 =0.00105 and trial 2 =0.00328/3=0.00109

molar mass of solid acid = mole = weight/molar mass so molar mass = weight/mole

trial 1=0.200/0.00105= 190.47

trial 2 =0.200/0.00109=183.48

average molar mass = (190.47+183.48)/2 ==>186.97

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