Titrations
1. A 50.00-ml NaOH sample of unknown concentration was titrated with 0.1274 M HCI. if 33.61 mL of the HCl solution were required to neutralize the NaOH sample, what is the molarity of the NaOH sample? Show the steps in your calculation
2. A 25.00-ml HCl sample of unknown concentration was titrated with 0.5631 M Al(OH)3. If 37.62 mL of the Al(OH) solution were required to neutralize the HCl sample, what is the molarity of the HCl sample? (Assume Al(OH)3 is acting as a strong base)
From the condition of equivalency we know,
where C is the molar concentrations of H+ (or actual acid) or, OH- (or actual base).
1. Here, Vacid = 33.61 mL , Cacid = 0.1274 M (because H+ is monoprotic) and Vbase = 50.00 mL
Thus,
So, the base concentration is 0.0856 M.
2. Here, Vbase = 37.62 mL, Cbase = 3 x 0.5631 M =1.6893 M (because actual base is OH- and it is supplying 3 moles of OH- per mole of Al(OH)3 ), Vacid = 25.00 mL
Thus,
So, acid concentration is 2.5420 M.
A 50.00-ml NaOH sample of unknown concentration was titrated with 0.1274 M HCI.
A 25.00 mL solution of H2SO4 of unknown concentration was titrated with a 3.15 M NaOH solution. What is the concentration of the H2SO4 solution if 36.01 mL of the NaOH solution is required neutralize the acid solution?
A 25.0 mL NaOH solution of unknown concentration was titrated with a 0.189 M HCl solution. 19.6 mL HCl was required to reach equivalence point. In a separate titration, a 10.0 mL H3PO4 solution was titrated with the same NaOH solution. This time, 34.9 mL NaOH was required to reach the equivalence point. What is the concentration of the H3PO4 solution?
a. Calculate the volume of 0.450 M Ba(OH), which will be needed to neutralize 46,00 mL of 0.252 M HCI. b. Find the molar concentration of a sulfuric acid solution, 35.00 mL of which neutralizes 25.00 mL of 0.320 M NaOH. (Careful! Sulfuric acid is diprotic!) c. Calculate the "molarity of water" H₂O in pure water at 30°C. (Hint: The density of water at 30°C is 0.9957 g/mL.] d. A 15.00 mL sample of a solution of H₂SO₄ of unknown concentration was titrated with...
Constant-boiling HCl can be used as a primary standard for acid-base titrations. A 50.00 mL sample of constant-boiling HCl with a concentration of 0.1225 M was collected and titrated to an end point with 35.60 mL of Ba(OH)2 solution. What is the molarity of the Ba(OH)2 solution?
Exercise 14.48- Enhanced - with Feedback Four solutions of unknown NAOH concentration are titrated with solutions of HCL The following table lists the volume of each unknown NAOH solution, the volume of HCl solution required to reach the equivalence point, and the concentration of each HCl solution. HCI (M) NAOH volume HCl volume (mL) (mL) 0.2099 8.00 8.97 11 14 0 1061 21.00 10 85 12.00 0.0889 35 18 29.00 0.1021 You may want to reference (Pages 494-497) Section 14.6...
Exercise 14.47 Four solutions of unknown HCl concentration are titrated with solutions of NaOH. The following table lists the volume of each unknown HCI solution, the volume of NaOH solution required to reach the equivalence point, and the concentration of each NaOH solution Part A Calculate the concentration (in M) of the unknown HCl solution in the first case. HCI Volume (ml) NaOH Volume (mL) (NaOH (M) 20.00 mL 27.44 mL 0.1191 M HCI Volume (mL) 20.00 mL 14.00 mL...
Four solutions of unknown HCl concentration are titrated with solutions of NaOH. The following table lists the volume of each unknown HCI solution, the volume of NaOH solution required to reach the equivalence point, and the concentration of each NaOH solution. Part BCalculate the concentration (in M) of the unknown HCl solution in the second case.
Exercise 14.47 Review Part Four solutions of unknown HCl concentration are titrated with solutions of NaOH. The following table lists the volume of each unknown HCI solution, the volume of NaOH solution required to reach the equivalence point, and the concentration of each NaOH solution. Calculate the concentration (in M) of the unknown HCl solution in the third case. HCI Volume (ml) NaOH Volume (mL) (NaOH (M) 24.00 mL 13.88 mL 0.1178 M Express your answer using four significant figures....
A 20.00 mL sample of HCl was titrated with the 0.022 M NaOH solution. To reach the endpoint required 23.72 ml of the NaOH. Calculate the molarity of the HCI. HCI + NaOH ----> NaCl + H2O Select one: O a. 0.026 M o b. 0.068 M o c. 0.039 M O d. 0.052 M
A 50.00 ml aliquot of 0.1000 M NaOH is titrated with 0.1000 M HCL. Calculate the pH of the solution after the addition of 0.00, 10.00, 25.00, 40.00, 45.00, 49.00, 50.00, 51.00, 55.00, and 60.00 ml of acid and prepare a titration curve from the data.