Write the full balanced equation for the reaction between acetic acid and sodium
ID: 1019865 • Letter: W
Question
Write the full balanced equation for the reaction between acetic acid and sodium hydroxides. A titration of 50.00 mL of acetic acid required 24.96 mL of 0.1056 M NaOH for complete neutralization. Calculate the molarity (moles/L) of the acetic acid. Show work & use proper Sig figs! At one point during the above titration (before the equivalence point), 10.52 mL of 0.1056 M NaOH, had been added to the 50.00 mL acetic acid and the pH was 4.47. Show all of your work for the following questions. Calculate the [H^+] at this point. Calculate the acetate ion concentration [A^-] that has been made when 10.52 mL of 0.1056 M NaOH was added. Calculate the acetic acid concentration [HA] when 10.52 mL of 0.1056 M NaOH was added to the 50.00 mL of acetic acid.Explanation / Answer
Solution:- (a) Acetic acid is CH3COOH and sodium hydroxide is NaOH. acetic acid is a weak acid where as sodium hydroxide is a strong base. Being an acid acetic acid gives H+ and NaOH gives OH-. These H+ and OH- combines to form H2O. The equation is written as...
CH3COOH(aq) + NaOH(aq) -------> CH3COONa(aq) + H2O(l)
(b) The equation used is, Ma. Va = Mb. Vb
Ma is molarity of acid and we are asked to find it out. Va is volume of acid given as 50.00 ml. Mb is molarity of base and given as 0.1056 M and Vb is volume of base given as 24.96 ml. all these given numbers have 4 sig figs so the answer also must have 4 sig figs. let's plug in the values and do the calculations.
Ma (50.00 ml) = 0.1056 M (24.96 ml)
divide both sides by 50.00 ml
Ma = 0.1056 M (24.96 ml) / 50.00 ml = 0.05272 M or 0.05272 mol/L
so, the molarity of the acid is 0.05272 M.
(c) We can calculate the H+ from given pH since, H+ = 10-pH = 10-4.47 = 3.39 x 10-5 M
(d) moles of NaOH added = 10.52 ml x (1L/1000ml) x (0.1056 mol/L) = 0.001111 mol
moles of acid added = 50.00 ml x (1L/1000ml) x (0.05272 mol/L) = 0.002636 mol
From balanced equation they react in 1:1 mol ratio. so, 0.001111 moles of each would to form 0.001111 moles of sodim acetate that is acetate ion since sodium acetate dissociates completely into Na+ and CH3COO-.
total volume is 50.00 ml + 10.52 ml = 60.52 ml = 0.06052 L
concentration of acetate ion[A-] = 0.001111 mol/0.06052 L = 0.01836 M
(e) from part d, we have added 0.001111 moles of NaOH and 0.002636 moles of acetic acid. excess moles of acetic acid = 0.002636 - 0.001111 = 0.001525 mol
Total volume = 0.06052 L
concentration of excess acetic acid, [HA] = 0.001525 mol/0.06052 L = 0.02520 M
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