Determine the pH of a 018MH_2CO_3 solution. Carbonic acid is a diprotic acid who

Question

Determine the pH of a 018MH_2CO_3 solution. Carbonic acid is a diprotic acid whose K_a1= 4.3 times 10^-7 and K_a2 = 56 times 10^-11. 10.44 5.50 11.00 3.56 4.31 Calculate the pH of a bugger that is 0.158 M HCIO and 0.099 M NaCIO. The K_a for HCIO is 2.9 time s 10^-8 6.67 733 377 646 A 1.0Lbuffer solution is 0250 M HC_2H_3O_2 and 0.050 M LIC_2H_3O_2. Which of the following actions will destroy the buffer? adding 0.050 moles of LiC_2H_3O_2 adding 0.050 moles of HC_2H_3O_2 adding 0.050 moles of NaOH adding 0.050 moles of HCI None of the above will destroy the buffer. Determine the concentration of Co_3^2- ions in a 0.18 M H_2CO_3 solution Carbonic acid is a diprotic acid whose K_al = -43 times 10^-7 and K_a2 = 5.6 times 10^_11. 32 times 10^-6 M 6.9 times 10^-8 M 5.6 times 10^-11 M 43 times 10^-3 M 2.8 times 10-4 M

Explanation / Answer

Answer (5)

H2CO3 is a diprotic acid (notice that there are two hydrogen atoms) so it has two Ka constants. Because constant Ka1 is so much greater than constant Ka2, you can assume that the concentration of the H3O+ ion is only attributable to Ka1.
First, write out the equilibrium expression, and use the ICE method:

H2CO3 + H2O <---> H3O+ + HCO3-
Initial: 0.18 M 0 0
Change: -x +x +x
Equilibrium: (0.18 - x) x x

Next, set up the Ka expression, and using your given value of Ka1, solve for x.

Ka = [H3O+][HCO3-] / [H2CO3] = 4.3 x 10-7

Water is not included in the expression because it is a liquid. Remember when solving for an equilibrium expression, concentrations of products go in the numerator, and concentrations of reactants go in the denominator.

Ka1 = x2 / (0.18-x) = 4.3 x 10-7

Assume that x is a small, and therefore negligible amount compared to 0.18.

Ka1 = x2 / 0.18 = 4.3 x 10-7

Rearrange to solve for x:

x = [(4.3 x 10-7)(0.18)]1/2 = 2.78 x 10-4 M = [H3O+]

To do this on a calculator: Multiply 4.3 x 10-7 by 0.18, then press the square root key on your calculator.

Now that you have the concentration of the H3O+ ion, you can find the pH of the H2CO3 solution. Remember:

pH = -log[H3O+]

Plug in the concentration you obtained:

pH of H2CO3 solution = -log[2.78 x 10-4 M] = 3.56

So the answer is (D) 3.56

Answer (6)

We knoe that

pH = pKa + log [ base / acid]

pH =- log 2.9 × 10 -8 + log [0.099 / 0.158]

pH = 7.538 -0.203 = 7.33

so the Answer is (B) 7.33

Answer (7)

A^- + HCl => HA + Cl^-
0.05..0.05.......0
0........0..........0.05
thus
a weak acid 0.05 M => pH = 3.02

so the answer is (D)adding 0.050 moles of HCl

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