A chemistry student weights out 0.291 g of citric acid (H3C6H5O7), a triprotic a

Question

A chemistry student weights out 0.291 g of citric acid (H3C6H5O7), a triprotic acid into a 250. mL volumetric flask and diluted to the mark with distilled water. He plans to titrate the acid with 0.0800 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 A chemistry student weights out 0.291 g of citric acid (H3C6H5O7), a triprotic acid into a 250. mL volumetric flask and diluted to the mark with distilled water. He plans to titrate the acid with 0.0800 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 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

Explanation / Answer

Molarity of citric acid = (mass of citric acid / molar mass of citric acid)*(1000 / volume of solution in L)

M1 = (0.291 / 192.) * (1000 / 250.)

M1 = 0.00606 M

Neutralisation reaction,

H3C6H5O7 (aq.) + 3 NaOH (aq.) -------------> Na3C6H5O7 (aq.) + 3 H2O(l)

Using neutralisation equation,

M1V1 / n1 = M2V2 / n2

0.00606 * 250. / 1 = 0.0800 * V2 / 3

V2 = 56.8 mL of NaOH is needed

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