What is th pH of a buffer solution made by dissolving 0.10 mol of formic acid, HCOOH, and 0.10 mol of sodium formate, HCOONa, in 1L of water?
What is the molarity of a solution made by dissolving 3.4g of Ba(OH)2 in enough water to make 450mL of solution? Assume that Ba(OH)2 ionizes completely in water to Ba2+ and OH- ions. What is the pH of the solution?
Use the Henderson-Hasselbalch equation for #1. For #2, you are to consider Ba(OH)2 a strong base. Post your work if you get stuck.
To find the pH of the buffer solution, we first need to determine the pH range of the buffer system. Formic acid (HCOOH) is a weak acid that ionizes in water, and its conjugate base, sodium formate (HCOONa), is a weak base that ionizes in water. The pH range of a buffer system is determined by the pKa value of the weak acid.
1. Calculate the pKa value of formic acid (HCOOH):
- The pKa value is often provided in a reference table, but if not, you can use an equation to calculate it. For formic acid, the pKa value is approximately 3.75.
2. Determine the pH range of the buffer system:
- The pH range of a buffer system is typically within one unit above and below the pKa value. In this case, the pH range will be from 2.75 to 4.75.
3. Calculate the concentrations of formic acid (HCOOH) and sodium formate (HCOONa) in the buffer solution:
- Since both substances are added in equal moles (0.10 mol), their concentrations in the buffer solution will be the same: 0.10 M.
4. Calculate the pH of the buffer solution:
- Since the concentration of HCOOH and HCOONa are the same, we can use the Henderson-Hasselbalch equation: pH = pKa + log [A-] / [HA]
- Substituting the values, we have: pH = 3.75 + log(0.10 / 0.10)
- Simplifying the equation, we find: pH = 3.75
Therefore, the pH of the buffer solution made by dissolving 0.10 mol of formic acid (HCOOH) and 0.10 mol of sodium formate (HCOONa) in 1L of water is approximately 3.75.
Now, let's move on to the second question.
To find the molarity (M) of the solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution in liters
1. Calculate the moles of Ba(OH)2:
- The molar mass of Ba(OH)2 is 171.34 g/mol.
- Given that we have 3.4 g of Ba(OH)2, we can calculate the moles using the formula: moles = mass / molar mass.
- Substituting the values, we have: moles = 3.4 g / 171.34 g/mol.
- Calculating the moles, we find: moles ≈ 0.01984 mol.
2. Calculate the molarity of the solution:
- The volume of the solution is given as 450 mL, which needs to be converted to liters. Since 1 liter is equal to 1000 mL, the volume of the solution is 450 mL / 1000 mL/L = 0.45 L.
- Now we can use the formula: Molarity (M) = moles of solute / volume of solution in liters.
- Substituting the values, we have: M = 0.01984 mol / 0.45 L.
- Calculating the molarity, we find: M ≈ 0.044 g/L.
To find the pH of the solution, we need to know the concentration of H+ ions, which is equal to the concentration of OH- ions since Ba(OH)2 ionizes completely.
3. Calculate the concentration of H+ ions:
- Since Ba(OH)2 ionizes completely, it forms 2 OH- ions for every 1 formula unit of Ba(OH)2.
- To find the concentration of OH- ions, we need to divide the moles of Ba(OH)2 by the volume of the solution and multiply by 2 (to account for the 2 OH- ions produced).
- Substituting the values, we have: concentration = (0.01984 mol / 0.45 L) × 2.
- Calculating the concentration, we find: concentration ≈ 0.088 M.
4. Calculate the pOH value:
- pOH is the negative logarithm (base 10) of the OH- concentration. We can calculate it using the formula: pOH = -log[OH-].
- Substituting the values, we have: pOH = -log(0.088).
- Calculating the pOH, we find: pOH ≈ 1.054.
5. Calculate the pH value:
- Since pH + pOH = 14 (for water at 25°C), we can calculate the pH using the formula: pH = 14 - pOH.
- Substituting the pOH value, we have: pH = 14 - 1.054.
- Calculating the pH, we find: pH ≈ 12.946.
Therefore, the pH of the solution made by dissolving 3.4g of Ba(OH)2 in enough water to make 450mL of solution is approximately 12.946.
To determine the pH of a buffer solution, you need to calculate the equilibrium concentration of the conjugate acid and conjugate base components of the buffer. The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffer solution:
pH = pKa + log([A-]/[HA])
where:
pH = the pH of the buffer solution
pKa = the acid dissociation constant of the weak acid component
[A-] = the concentration of the conjugate base component
[HA] = the concentration of the weak acid component
For the first question:
1. Calculate the pKa of formic acid (HCOOH). Since the question doesn't provide the pKa value, you need to look it up. The pKa of formic acid is approximately 3.75.
2. Calculate the concentration of the conjugate base component ([A-]). Since it is a 1L solution containing 0.10 mol of sodium formate (HCOONa), the concentration of HCOO- is 0.10 M (assuming sodium ion, Na+, doesn't contribute to the pH).
3. Calculate the concentration of the weak acid component ([HA]). It is also 0.10 M since the amount of formic acid (HCOOH) is given as 0.10 mol in 1L of water.
4. Substitute the values into the Henderson-Hasselbalch equation:
pH = 3.75 + log(0.10/0.10)
pH = 3.75 + log(1)
pH = 3.75 + 0
pH = 3.75
Therefore, the pH of the buffer solution made by dissolving 0.10 mol of formic acid and 0.10 mol of sodium formate in 1L of water is 3.75.
For the second question:
1. Calculate the molarity (M) of the Ba(OH)2 solution by using the formula:
Molarity (M) = moles of solute / volume of solution in liters
Given:
Mass of Ba(OH)2 = 3.4 g
Volume of solution = 450 mL = 0.450 L
M = (3.4 g / (Ba(OH)2 molar mass) / 0.450 L)
Since the molar mass of Ba(OH)2 = (137 + (2 * 16) + (2 * 1)) = 171 g/mol:
M = (3.4 g / 171 g/mol) / 0.450 L
M ≈ 0.040 M
Therefore, the molarity of the Ba(OH)2 solution is approximately 0.040 M.
2. Since Ba(OH)2 is a strong base, it dissociates completely in water to form Ba2+ and OH- ions. Each Ba(OH)2 molecule produces two OH- ions, so the concentration of OH- in the solution is 2*0.040 M = 0.080 M.
3. To calculate the pH of the solution, we need to find the concentration of H+ ions, which can be determined using the equation:
pOH = -log10[OH-]
Given the concentration of OH- as 0.080 M:
pOH = -log10(0.080)
Using the property that pOH + pH = 14, we can calculate the pH:
pH = 14 - pOH
pH = 14 - (-log10(0.080))
Now, calculate:
pH ≈ 14 - (-1.097)
pH ≈ 15.097
Therefore, the pH of the solution made by dissolving 3.4g of Ba(OH)2 in enough water to make 450mL of solution is approximately 15.097.