Calculate the pH of a solution containing a salt AcNa derived from a strong base (NaOH) and a weak acid (AcH), like CH3COONa, KF, NaNO2 and so on. In this case two processes have to be considered:
1. Dissociation of the salt, within the assumption that the salt is a strong electrolyte:
AcNa → Ac- + Na+, for instance: CH3COONa → CH3COO- + Na+
2. the hydrolysis of the water
Ac- + H2O ⇌ AcH + OH-, for example: CH3COO- + H2O ⇌ CH3COOH + OH-
knowing that the initial concentration of the salt is 1.7 M and the acid dissociation constant, Ka , is 2.23e-10.
pH=?
Calculate the pH of a solution containing a salt AcNa derived from a strong base (NaOH)...
A solution contains 1M weak acid (AH), 100mM salt ANa (strong electrolyte), and 1mM HCl(strong acid). The pH of the solution is 2. Determine the pKa of the weak acid AH. The dissociation processes are as follows: AH<-> A- + H+, ANa-> A- + Na+, HCl-> H+ + Cl-
Since only salt NaA and H2O present in the solution so H+ ion comes from hydrolysis of NaA to give NaOH and HA and dissociation of H2O. Hence we calculate pH by formula for salt formed by weak acid and strong base.
Salt of a Weak Base and a Strong Acid. pH of Solution. Calculate the pH of a 1.19 M aqueous solution of triethylamine hydrochloride ((C2H5)3NHCI) (For triethylamine, (C2H5)3N, Kb = 4.00x 10-4.) Give two decimal places in your answer.
1) Calculate the pH of the 1L buffer composed of 500 mL of 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of HCl is added (Ka HC2H3O2 = 1.75 x 10-5). Report your answer to the hundredths place. 2) Calculate the pH of the 1L buffer composed of 500 mL 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of NaOH is added (Ka HC2H3O2 = 1.75...
8. What is the pH of a 5.00 x 102 M Ba(OH)2(aq) solution at room temperature? a. 1.00 b. 1.30 c. 12.02 d. 12.70 e. 13.00 9. Assuming equal concentrations of conjugate acid and base, which one of the following mixtures is suitable for making a buffer solution with an optimum pH of 4.6-4.8? a. CH3COONa/CH3COOH (K = 1.8 x 105) b. NH/NHACI (K = 5.6 x 10-1) c. NaOCI/HOCI (Ka = 3.2 x 108) d. NaNO2/HNO2 (Ka = 4.5...
Salt Hydrolysis: Hydrolysis at the Endpoint of a Titration Consider the following equation: NH3 (aq) + HBr (aq) → NH4Br (aq) + H2O (l) Here the neutralization reaction is creating ammonium bromide. If you add enough strong acid to react with all of the weak base, all that will remain will be ammonium bromide and water. The conjugate acid from this salt will hydrolyze in solution to determine the pH at this point. Weak Acid Ka Weak Base Kb CH3COOH...
a. The pH of a strong base solution is eaqsy to calculate because the salt molecule dissociates 100%. So if 2.5 moles of NaOH are dissolved in water, that solution will have how many moles of OH- ions? b. If 0.03 moles of Ca(OH)2 are dissolved in water, that solution will have many moles of OH- ions?
1)A buffer solution was prepared by dissolving 3.95 g of sodium nitrite, NaNO2, in 150 mL of 0.200 M nitrous acid, HNO2. (Ka = 4.5 x 10-5). What is the pH of the buffer? 2) Consider a buffer CH3COOH/CH3COO–. Fill in the blank using the numbers corresponding to each species. CH3COOH CH3COO– OH– H+ Na+ H2O 1 2 3 4 5 6 Please note that choices can be used more than once. The basic component of the buffer is ....
1. Determine the Ka value for an acid that is 0.294 M in a solution that has a pH of 2.80. A 49 x 10-14 B. 23 x 109 C. 14 x 10-5 D. 20 x 10-5 OE 3.7x 10-7 2. When titrating 25.0 mL of 0.080 M HCIO with 0.060 M NaOH, calibrate the pH after adding 25.0 mL of 0.050 M NaOH (pKa = 7.55) A. 7.43 b.7.55 C.7.77 D.10.08 E 1248 3. The correct expression for the...
Weak Acid-Strong Base Titration Date Nanse PRE-LAB OUESTIONS 1 Calculate the molarity of a NaOH solution that was used to titrate 1.2 g of potassium acid phthalate if 37.50 ml of the base were reauired to get to the end point of the titration. 2 It takes 12.45 ml. of a 0.500 M NaOH solution larity of the acid to titrate 30.0 mL of acetic acid. What is the mo- 3. Using the titration curve below, calculate the K, of...