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Levart [38]
2 years ago
6

calculate the ph of a 0.020 m carbonic acid solution, h2co3(aq), that has the stepwise dissociation constants ka1

Chemistry
1 answer:
fgiga [73]2 years ago
5 0

The calculated pH is 3.79. therefore, the solution is acidic.

No, carbonic acid is not a strong acid. H2CO3 is a weak acid that dissociates into a proton (H+ cation) and a bicarbonate ion (HCO3- anion). This compound only partly dissociates in aqueous solutions.

H2 CO3    =    H (+) + HCO3(-)          Ka1 = 4.3 * 10^ -7

   0.06 - x              x          x

Ka1 = x^2 / (0.06 - x) = 4.3 * 10^ - 7

A low Ka => x << 0.06 => 0.06 -x ≈ 0.06

=> Ka1 ≈ x^2 / 0.06 => x^2 ≈ 0.06 * Ka1 = 0.06 * 4.3 * 10^-7

=> x ≈ √ [ 2.58 * 10 ^ -8] = 1.606 * 10^ - 4 = 0.0001606

Second dissociation

HCO3(-)    =   H (+) + CO3(2-)         Ka2 = 5.6 * 10^ - 11

0.0001606 - y            y             y

Ka2 ≈ y^2 / 0.0001606 => y = √ [0.0001606 * 5.6* 10^ -11]

y = 9.48 * 10^ -8

An acidic solution has a high concentration of hydrogen ions (H +start superscript, plus, end superscript), greater than that of pure water.

[H+] = x + y = 1.607 * 10^ -4

pH = - log [H+] = 3.79

Learn more about concentration here-

brainly.com/question/10725862

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Since the temperature is a constant, we can use Boyle's law to solve this.<span>

<span>Boyle' law says "at a constant temperature, the pressure of a fixed amount of an ideal gas is inversely proportional to its volume.

P α 1/V               </span>⇒                 PV = k (constant)<span>

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<span>Here, we assume that the </span>gas in the balloon is an ideal gas. 

We can use Boyle's law for these two situations as,
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By substitution,
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8 0
3 years ago
One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO: 4NH3 (g) 5O2 (g
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Answer:

O2 is the limiting reactant.

Explanation:

Step 1: Data given

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Step 2: The balanced equation

4NH3(g) +  5O2 (g) → 4NO(g) +  6H2O (g)

Step 3: calculate moles NH3

Moles NH3 = mass NH3 / molar mass NH3

Moles NH3 = 2.00 grams / 17.03 g/mol

Moles NH3 = 0.117 moles

Step 4: Calculate moles O2

Moles O2 = mass / molar mass O2

Moles O2 = 2.50 grams / 32 g/mol

Moles O2 = 0.0781 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed. (0.0781 moles). NH3 is in excess. There will react 4/5 * 0.0781 moles =  0.0625 moles

There will remain 0.117 - 0.0625 = 0.0545 moles NH3

O2 is the limiting reactant.

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