[Ipassedthisclass]

1. The shell of a crustacean is primarily made of Calcium carbonate. As ocean acidification becomes more of a problem in our climate, crustaceans are suffering from habitat destruction.
H2O (l) + HCO3- ←→ H3O+ (aq) + CO32- (aq)           Ka = 4.7 x 10-11
You take a 1.0-Liter sample of ocean water and determine that there are 8.127 x 1017 molecules of HCO3-.  What will be the pH of the ocean based on this sample?

2. Sodium Nitrate mixes with potassium carbonate in a solution of water. Write out the ionic equation.

3. Sulfurous acid is added to water. Write out the ionic equation.

[Aldebaran]

Rules of this forum require that you demonstrate your own attempts at the questions first rather than just asking someone to do your homework for you.

[Vidya]

1. The shell of a crustacean is primarily made of Calcium carbonate. As ocean acidification becomes more of a problem in our climate, crustaceans are suffering from habitat destruction.
H2O (l) + HCO3- ←→ H3O+ (aq) + CO32- (aq)           Ka = 4.7 x 10-11
You take a 1.0-Liter sample of ocean water and determine that there are 8.127 x 1017 molecules of HCO3-.  What will be the pH of the ocean based on this sample?

You need to review how to get pH of weak acids using RICE table.

. Sodium Nitrate mixes with potassium carbonate in a solution of water. Write out the ionic equation.

3. Sulfurous acid is added to water. Write out the ionic equation.

Start with strong electrolytes and weak electrolytes .

[Blackwhite]

When a strong acid and a strong base are mixed, they react according to the following net-ionic equation: H₃O⁺(aq) + OH⁻(aq) → 2H₂O(l). If either the acid or the base is in excess, the pH of the resulting solution can be determined from the concentration of excess reactant.

[Piers Simon]

Let x be the concentration of H3O+ (mol/L) formed when HCO3- reacts. Since H2O is present in excess, we can assume that the initial concentration of HCO3- is equal to the concentration formed:

[HCO3-] = x (mol/L)
[H3O+] = x (mol/L)

The equation for the equilibrium constant (Ka) is given as:

Ka = [H3O+][CO32-] / [HCO3-]

Given Ka = 4.7 x 10^-11 and [HCO3-] = 8.127 x 10^17 molecules in 1.0 liter (which is equivalent to 8.127 x 10^-8 mol/L):

4.7 x 10^-11 = (x)(x) / (8.127 x 10^-8)

Solving for x:

x^2 = 4.7 x 10^-11 * 8.127 x 10^-8
x^2 = 3.82969 x 10^-18
x ≈ 1.956 x 10^-9 mol/L

Now, to find the pH, we can use the equation:

pH = -log[H3O+]

pH ≈ -log(1.956 x 10^-9)
pH ≈ 8.71

So, the pH of the ocean based on this sample is approximately 8.71.

Ionic equation for the reaction of Sodium Nitrate (NaNO3) with Potassium Carbonate (K2CO3) in water:
NaNO3(aq) + K2CO3(aq) → 2KNO3(aq) + Na2CO3(aq)

Ionic equation for the addition of Sulfurous Acid (H2SO3) to water:
H2SO3(aq) → H+(aq) + HSO3-(aq)

[Borek]

So, the pH of the ocean based on this sample is approximately 8.71.

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