Acids and bases
• What is an acid?
• A 500cm3 solution containing 1.9g of a weak acid HA has a pH of 3.50. Calculate the molar mass of the acid, given that it has a Ka of 2.0 × 10-6moldm-3
• Calculate the pH of the following:
• 0.002moldm-3 NaOH
• 0.30 moldm-3 NH4Br
• 0.001moldm-3 H2SO4
You can get the definition of an acid from you text/notes/google.
For the HA problem. For pH = 3.50 convert to (H^+). I obtained approx 3E-4 but you need to confirm since that's an estimate.
.......HA ==> H^+ + A^-
E......x....3E-4...3E-4
Substitute the E line into the Ka expression and solve for (HA).
Then (HA) = M = mols/L. YOu have M and L, solve for mols.
Then mols = grams/molar mass. You have mols and grams, solve fo rmolar mass. Remember to use the real value for (H^+) and not my estimate.
By the way, the Ka value has no units and don't let anyone tell you they do. When they are used, most of the time they are called just units, but often they are called "tentative units". "temporary units", "trial units", or
Acids and bases:
• An acid is a chemical substance that donates protons (H+ ions) or accepts a pair of electrons in a chemical reaction. Acids typically have a sour taste and can react with metals, carbonates, and bases to produce salts and water.
Calculating the molar mass of acid HA:
To calculate the molar mass of the acid HA, we can start with the given information:
- Solution volume: 500 cm^3 (which is equal to 0.5 dm^3 since 1 dm^3 = 1000 cm^3)
- Mass of weak acid HA: 1.9 g
- pH of the solution: 3.50
- Ka value of the acid: 2.0 × 10^-6 moldm^-3
The pH of a solution can be related to the concentration of H+ ions using the formula: pH = -log[H+]. To find the concentration of H+ ions, we can rearrange the formula: [H+] = 10^(-pH).
Since the concentration of the weak acid HA and the concentration of its dissociated species (H+ and A-) are the same when the acid is weak, we can assume that [HA] = [H+].
Using the Ka expression for the acid dissociation: Ka = [H+][A-] / [HA], we can substitute [HA] = [H+] into the equation.
Ka = [H+][H+] / [HA]
2.0 × 10^-6 = [H+]^2 / [HA]
Since we have the pH value (which corresponds to [H+]), we can substitute it into the equation:
2.0 × 10^-6 = ([HA]/0.5)^2
Solving for [HA], we get: [HA] = √(2.0 × 10^-6 × 0.5^2) = 5 × 10^-4 mol/dm^3
Now, we can calculate the number of moles of the weak acid using the concentration and volume of the solution:
Number of moles = concentration × volume = 5 × 10^-4 mol/dm^3 × 0.5 dm^3 = 2.5 × 10^-4 moles
Finally, we can calculate the molar mass of the acid HA using the formula:
Molar mass = (mass of acid) / (number of moles)
Molar mass = 1.9 g / 2.5 × 10^-4 moles = 7600 g/mol
Therefore, the molar mass of the acid HA is 7600 g/mol.
Calculating the pH of the given solutions:
To calculate the pH of each solution, we need to consider the concentration or molarity of H+ or OH- ions.
1. 0.002 moldm^-3 NaOH:
NaOH is a strong base that dissociates completely in water to produce OH- ions. Since the concentration of OH- ions is the same as the concentration of NaOH, the concentration of OH- is 0.002 moldm^-3. To find the pOH, we can use the equation pOH = -log[OH-]. Then, the pH can be calculated using the equation pH + pOH = 14.
pOH = -log(0.002) = 2.7
pH = 14 - pOH = 14 - 2.7 = 11.3
Therefore, the pH of a 0.002 moldm^-3 NaOH solution is 11.3.
2. 0.30 moldm^-3 NH4Br:
To calculate the pH of this solution, we need to consider the behavior of the ammonium ion (NH4+). NH4+ is a weak acid that can donate a proton in water. Since it is a weak acid, we can assume that only a fraction of NH4+ dissociates. However, NH4Br is a salt and can also react with water to form OH- ions.
First, we need to calculate the concentration of NH4+ ions. Since it is a 1:1 salt, the concentration of NH4+ is the same as the concentration of NH4Br, which is 0.30 moldm^-3.
Next, we need to consider the hydrolysis of NH4+, which will produce some OH- ions. The hydrolysis reaction is NH4+ + H2O ⇌ NH3 + H3O+. However, ammonia (NH3) is a weak base, and the concentration of H3O+ is negligible considering its weak acid properties. Therefore, we can ignore the contribution of H3O+ and consider OH- as the dominant species.
Since NH4+ reacts with water to produce OH- ions, the concentration of OH- ions will increase. To calculate the pOH, we can use the equation pOH = -log[OH-]. Then, the pH can be calculated using the equation pH + pOH = 14.
pOH = -log(0.30) = 0.52
pH = 14 - pOH = 14 - 0.52 = 13.48
Therefore, the pH of a 0.30 moldm^-3 NH4Br solution is 13.48.
3. 0.001 moldm^-3 H2SO4:
H2SO4 is a strong acid that dissociates completely in water to produce H+ ions. Therefore, the concentration of H+ ions is 0.001 moldm^-3. To find the pH, we can use the equation pH = -log[H+].
pH = -log(0.001) = 3
Therefore, the pH of a 0.001 moldm^-3 H2SO4 solution is 3.