Question

O Even to Last Response 2. -/0.1 points 0/4 Submissions Used Assume you dissolve 0.235 g of the weak acid benzoic acid, C.HCO2H, in enough water to make 7.00 x 102mL of solution and then titrate the solution with 0.138 M NaOH. CHECO2H(aq) + OH(aq) + CHCO2 (aq) + H2010 What are the concentrations of the following ions at the equivalence point? Na*, H30, OH CH5CO2 M Na MHz0+ Мон" M CHECO2 What is the pH of the solution? 3. 0.1/0.1 points Previous Answers 1/4 Submissions Used Construct a rough plot of pH versus volume of added base for the titration of 50 mL of 0.055 M HCN with 0.075 M NaOH, K, - 4.0-10-10 for HCN a Search or type URL 8 9 0 7 4 5 6
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Answer 1

Males of benzoic acid - may 0.2359 53 = 0.00192 molaumas 122. 12g Amal mot Concentration of benzoic acid = 0.00192X/0 0.0027M 72 at equivalence faint moles Noon needed= 0.00192 mol so. volume of Naon = 0.00192 mol 0.0139 0.138M Total volume = 0.7+0.0139 = 0.7139L at equivalence peint. moles of Cong(0o = 0.00192 Contentration of CHOCOo= 0.00192 mol = 2.7 mm 0.7139 L Colo Coo- +120 - Cote Coon + on Kb = kw. _ 10-14 = 1:6810-10 Ка с. 3X (о- ke x² rikollel 0.00268 -- 0.00268->20-00268 1.6810-10 = x2 0.00268 X = 16X16X0.00268= 0655X/0-6

0.655x100M x= [on] = [ch₂ Cook] [ut] 10-14 = 1053X10-8 0.655x10-6 Pu= -log[ht] Py = -log (1.53 X108) PH 7.815 Answers a) [130"] = 1.53X10-8 M b) [ok] = 0.655X106m C) [ Coh,600] = 2069 X10-3M d) Nat] = 2.69 X10 3M Thank you, May this help you .