Q. The pH of a solution of 1.1M H2A (Ka1 = 1.0 x 10^-6 an Ka2 = 1.0 x 10^-10) is:
a. 10.00
b. 5.96
c. 11.02
d. 2.98
e. None of these
The acid H2A is a two-protonically weak acid with Ka1 = 1.0 * 10^-2 and Ka2 = 1.0 * 10^-6 . Sketch concentration of H2A, HA- , and A2- towards the pH of the solution, starting from 1.0 M H2A. Explain also why the concentrations vary as they do.
Question 4 (2 points) The pH of a solution of 1.1 MH2A (Kal = 1.0 x 10-6 and K 2 is 1.0 x 10-10) is: 0 10.00 5.96 C-11.02 2.98 none of these
Calculate the pH when 200.0 mL of a 1.00 M solution of H2A (Ka1 = 1.0 x 10–6, Ka2 = 1.0 x 10–10) is titrated with the following volumes of 1.00 M NaOH. 250.0 mL of 1.00 M NaOH Group of answer choices
For the diprotic weak acid H2A, Ka1=2.3×10−6 and Ka2=7.2×10−9. What is the pH of a 0.0400 M solution of H2A? pH= What are the equilibrium concentrations of H2A and A2− in this solution? [H2A]= M [A2−]= M
For the diprotic weak acid H2A, Ka1 = 4.0 × 10-6 and Ka2 = 6.1 × 10-9. What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
For the diprotic weak acid H2A, Ka1 = 3.8 × 10-6 and Ka2 = 5.8 × 10-9. What is the pH of a 0.0700 M solution of H2A? What are the equilibrium concentrations of H2A and A2- in this solution?
For the diprotic weak acid H2A, Ka1 = 3.8 × 10-6 and Ka2 = 6.9 × 10-9. What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
For the diprotic weak acid H2A, Ka1 = 2.8 × 10-6 and Ka2 = 8.2 × 10-9. What is the pH of a 0.0500 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
For the diprotic weak acid H2A, Ka1 = 3.5 × 10-6 and Ka2 = 6.2 × 10-9. What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
For the diprotic weak acid H2A, Ka1 = 3.6 × 10-6 and Ka2 = 6.7 × 10-9. What is the pH of a 0.0500 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?