of 204.23 g/mol to A student uses a weak monoprotic carboxylic acid with a molec

Question

of 204.23 g/mol to A student uses a weak monoprotic carboxylic acid with a molecular weight standardize an NaOH solution. He dissolved 0.500 g of the weak acid in 100.0 mL of DI water and adds phenolphthalein indicator. This solution is then titrated with NaOH until the solution turns pink. The initial reading of the buret containing the NaOH was 50.0 mL. A) (3 POINTS) How many moles of weak acid were in the flask? 50m204.23g B) (3 POINTS) How many moles of NaOH were used to reach the equivalence point? C) (4 POINTS) The final buret reading was 36.4 mL. Calculate the molarity of the NaOH.

Explanation / Answer

A: Moles of weak acid in the flask = mass / molar mass = 0.500 g / 204.23 g/mol = 0.00245 mol

B: The neutralization reaction is

CH3COOH + NaOH ----- > CH3COONa + H2O

1 mol, -------- 1 mol ---------- 1 mol

Hence moles of NaOH used to reach equilivance point = moles of carboxylic acid = 0.00245 mol

(C): Volume of NaOH used = 50.0 mL - 36.4 mL = 13.6 mL = 0.0136 L

moles of NaOH used = 0.00245 mol = MxV = M x 0.0136 L

=> M = 0.00245 mol / 0.0136 L = 0.180 M (answer)

Q.2: (A):Moles of acid in 25 mL = moles of NaOH used = MxV = 0.10 mol/L x (19.9 - 0.4) mL x (1L / 1000 mL)

= 0.00195 mol

mass of acid in 25 mL solution = (4.10 g/1000mL ) x 25 mL = 0.1025 g

=> moles of acid = 0.00195 mol = 0.1025 g / molar mass

=> molar mass = 0.1025 g / 0.00195 = 52.6 g/mol

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