Question

A chemistry student weighs out 0.115 g of sulfurous acid (H,SO,), a diprotic acid, into a 250 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1500 M NaOH solution Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits x 5 ?

Answer 1

No. of moles of sulfurous and = 0.1159 -0.0014 moles. 82 g/mol Molarity of sulfurous acid solü. = No. of moles Vot of soli in litre Molarity 0.0014 moles = 0.0056M 0.250L Chemical equr :- H₂SO3 + 2 NaOH -- Na, SO3 + 2 H₂O Chemical equ? shows that I moles of NaOH are needed for neutralisation if I mole of Haslz (diprofic acid). Applying conc. equl fir final esuivalence point of this diprotic acid, Acid 2 (0.0056M) x 250 mL = 0.1500 MX V No = 480 m = 18.67 me 0.1500 ry Hence the student requires [18.67 mL of 0.1500 M NaOH solu". to reach the final equivalence point. Rate Please 18 .67 ML