9. Add about 20 mL of water to the the endpoint, une poin flask to 0. Use techni
ID: 1061128 • Letter: 9
Question
9. Add about 20 mL of water to the the endpoint, une poin flask to 0. Use techniques demonstrated your instructor to titrate the acid seconds. with WORAN by and pink for at least 30 the flask which line pink mL). Rinse the sides of the volume reading on the beyond the bottom (25 mL or 50 neut the occasionally to make certain all the acid being poured down the scale) Pink solutions may be 11. Rinse the Erlenmeyer flask and repeat steps 8-9 two more times (three trials total). If the burette still than halffull, the next on can start where the last one ended. Otherwise refill the burette with NaOH solution. If the first required less than 10 mL of base, acid in subsequent titrations so that they require at least 10 mL to reach the end point. 12. Unless you will go right on to titrate vinegar, d excess NaoH solution back into the storage bottle, seal the bottle tightly and store it for the next lab period. Thoroughly rinse the burette, inc uding the tip, with tap water and drain it completely before returning it. C Data for standardization: Table I Trial 3 Trial 2 Trial 1 0.5083 Mass KHP (s) U50159 40 mL Final burette reading Initial burette reading Volume of base used 13.20 mL 13.00 ML subeact initial from final) D. Calculations Write a molecular equation for the reaction above: Write the equation as a net ionic equation (note that HCsHro is a weak acid.)Explanation / Answer
Acid-base titration
10. moles of H2SO4 present = 0.5544 M x 25 ml = 13.86 mmol
moles of KOH needed = 2 x 13.86 mmol = 27.72 mmol
ml of KOH needed = 27.72 mmol/0.8561 M = 32.4 ml
11. It is alright to add water during the titration of KHP and KOH. The KOH added to a solution of KHP in a flask. Moles of KHP remained constant and so any amount of water added to the flask would not change this moles of KHP present and it would take the same volume of KOH to neutralise the acid in solution.
12. Volume of HNO3 required to neutralize 2.02 g Ca(OH)2
moles of Ca(OH)2 = 2.02 g/74.093 g/mol = 0.027 moles
1 mole of Ca(OH)2 (dibasic) would take 2 moles of HNO3 (monoprotic acid) for complete neutralization
ml of HNO3 needed = 2 x 0.027 mol/0.2269 M = 240.63 ml
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