2. It requires 27.5 mLofa NaOH solution to titrate 0.214 g of H,C,04 (MM=90.0).
ID: 1084200 • Letter: 2
Question
2. It requires 27.5 mLofa NaOH solution to titrate 0.214 g of H,C,04 (MM=90.0). What is the molarity of the NaOH solution? 3. It requires 42.5 mL of a NaOH solution to titrate 25.0 mL of a 0.192 M HCI solution. If 32.4 mL of the NaOH solution is required to titrate a solution containing 0.410 g of an unknown acid, what is the molar mass of the monoprotic unknown acid? 4. It requires 17.5 mL of a Sr(OH2 solution to titrate a solution containing 0.613 g KHP (MM = 204.2 g/mol). If 31.4 mL of the Sr(OH)2 solution are required to titrate 7.04 g of an impure HC2H3O2 (MM = 60.0) solution, what is the % of HC2H,02 in the solution? A 0.104 M NaOH solution is used to titrate a solution containing 0.124 g KHP (MM = 204.2) and 0.140 g H2C204 (MM = 90.0). What volume (in mL) of NaOH is required? . SWERS 2. 0.173 M 3.112 g/eq 4.459% 5.35.8 mL 58.3 mLExplanation / Answer
Oxalic acid is a diprotic acid (each mole of acid gives two moles of protons) while sodium hydroxide is a monobasic base (one mole of hydroxide per mole of NaOH). Thus by balancing the neutralization reaction we get the reaction as: 2NaOH + H2C2O4 -----> Na2C2O4 + 2H2O which helps one state that for every mole of oxalic acid neutralized, two moles of sodium hydroxide is required.
From the molar mass of oxalic acid to be 90, the no.of moles of the acid in 0.214g of it will be 0.214g/(90g/mol) = 2.3778x10-3mol implying that 2*2.3778x10-3mol = 4.7556x10-3mol of NaOH is required. Therefore it can be seen that 4.7556x10-3mol of NaOH was present in 27.5mL of the base solution. Thus the molarity of the solution which is defined as the no.of moles of substance per liter of solution is calculated as Molarity = Moles of NaOH*(27.5/1000) = 1.3078x10-4M.
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