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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Ipassedthisclass on May 23, 2023, 12:06:56 AM
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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.
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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.
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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 .
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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.
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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)
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So, the pH of the ocean based on this sample is approximately 8.71.
Please read the forum rules (https://www.chemicalforums.com/index.php?topic=65859.0). You clicked "I did" while registering, but apparently you lied.