Many homeowners treat their lawns with CaCO3(s) to reduce the acidity of the soil. Write a net ionic equation for the reaction o
1 answer:
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Answer:
In an acid-base equilibrium, acid becomes a conjugate base and base becomes a conjugate acid.
Explanation:
Let's remember the Bronsted-Lowry theory to answer this specific question. According to the theory, acid is a proton donor, while a base is a proton acceptor.
Consider an acid in a form HA (aq) and base in a form of B (aq). Since acid is a proton donor, it will donate its hydrogen ion to the base, B. The resultant products would be (aq) and
(aq).
Remember that an acid-base reaction is an equilibrium reaction. This means we may also look at this proton transfer reaction from the product side towards the reactants. Summarizing what has been said, we may write the equilibrium as:
⇄
Now acid, HA, donates a proton to become a conjugate base. The conjugate base, if we look from the reverse equation side, is actually a base, since it can accept a proton to become HA. Similarly, B accepts a proton to become a conjugate acid. Looking from the reverse reaction, it can now donate a proton, so in reality we can consider it a base.
To summarize, your logic is correct.
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.