Calculate the pH of each of the following diprotic acid solutions is 25C, ignoring second deprotonations only when that approximation is justified.
a) 0.10 M H2Saq
b) 0.15 M H2C4H4O6aq
c) 1.1E-3 M H2TeO4aq Ka1=2.1E-8, Ka2=6.5E-12
Please don't hestitate to drop some knowledge on me :)


Calculate the pH of each of the following diprotic acid solutions is 25C, ignoring second deprotonations...
A diprotic acid, H2A, has acid dissociation constants of
Ka1=1.01×10−4 and Ka2=4.08×10−12. Calculate the pH and molar
concentrations of H2A, HA−, and A2−at equilibrium for each of the
solutions.
A diprotic acid, H, A, has acid dissociation constants of Kal = 1.01 x 104 and K22 = 4.08 x 10-12. Calculate the pH and molar concentrations of H, A, HA, and A? at equilibrium for each of the solutions. A 0.176 M solution of H, A. pH= pH = 1...
Consider a diprotic acid, H2A, with the following Ka values. Ka1 = 0.01 Ka2 = 0.008 If you have a 0.01 M solution of H2A, what is [H3O+] and the pH ? Hint: The Ka values are too close together to ignore the second equilibrium.
Calculate the pH of each of the following aqueous solutions at 25 °C. (Please help me solve #4 with your work, and check my answers for #1-3, if incorrect please show me why, Thank you) 1) 0.65 M boric acid (B(OH)3, Ka = 7.3 x 10-10) (pH=4.66) 2) 3.15 M ammonia (NH3, Kb = 1.76 x 10-5) (pH= 4.37) 3) 0.82 M benzoic acid (C6H5COOH, Ka = 6.3 x 10-5) (pH=2.14) 4) 0.100 M H3AsO4 (Ka1 = 2.5 x 10-4,...
Consider the following diprotic acid. H2AH20 Ho+HA Kal-5.0x105 +A2- The initial concentration of H2A = 1.0 × 10-5 M. HA" + H20 H30" Kai = 4.5 x 10-5 Lal 1. Since the Ka and Ka2 values are very close, treat both reactions simultaneously. (a) Determine the H3+ at equilibrium 2 Now, as a comparison, ignore the second reaction (a) Determine the [H0. 3. Is the approximation of ignoring the second reaction valid?
A diprotic acid, H2A, has acid dissociation constants of ?a1=3.69×10−4 and ?a2=4.08×10−12. Calculate the pH and molar concentrations of H2A, HA−, and A2− at equilibrium for each of the solutions. A 0.102 M solution of H2A. A 0.102 M solution of NaHA. A 0.102 M solution of Na2A
Calculate the pH of each of the following strong acid
solutions.
Calculate the pH of each of the following strong acid solutions. (a) 0.00696 M HI pH = (b) 0.719 g of HCl in 29.0 L of solution pH = (c) 49.0 mL of 2.10 M HI diluted to 3.40 L pH = (d) a mixture formed by adding 87.0 mL of 0.000520 M HI to 47.0 mL of 0.000860 M HCI pH =
Calculate the pH of the following aqueous solutions. Constants 0.15 M HC2H302 (acetic acid) R=0.08206 L.atm/molek R=7.314 J molek Ko '(7273.15 a. Show Chemical equation b. 0.10 M NaC3H5O2 (sodium propionate) c. 0.010 M HBr (hydrobromic acid)
1. Using the given information answer the following questions with corresponding answers. (a) Serine is a diprotic acid (Ka1 = 6.17 × 10-3 and Ka2 = 7.08 × 10-10) that can have three different forms in solution: H2S+, HS, and S- (Note: S is not sulfur in these formulas). What would be the concentration of H2S+ in the solution of 0.105M Na+S-? Group of answer choices 8.29 × 10-8 M 1.62 × 10-12 M 9.62 × 10-9 M 4.15 ×...
1. Complete an ICE table and calculate the pH for the following solutions. a. 0.50 M HC,H,O2 (benzoic acid, Ka = 6.5 x 10) b. 0.10 M CH3NH2 (methylamine, Kb 4.4 x 10) c. 0.50 M NaC7H,O2 (Sodium benzoate) d. 0.10 M CH3NH3CI (methylammonium choride) e. 0.10 M Na2SO4 (sodium sulfate; Ka2 for H2SO4 0.012) 2. What is an amphoteric species? Provide an example of an amphoteric species and write balanced equations that show why it's amphoteric.
Calculate the pH of each of the following solutions of a strong acid in water. a. 0,010 M HCI pH = 6.3.6 M HC104 pH = c. 5.6 x 10-11 M HI pH =