3. Determine the pH if 0.02 mol of NaOH is added to 1.0L of the buffer described in question #1.
question1: 1. How many grams of NaHCO3 should be added to one liter of 0.100 M H2CO3 (Ka = 4.2 x 10-7) to prepare a buffer with pH = 7.00?
1st: equation: H2CO3+NaOH<-->H2O+HCO3
2ndICE For Initial: H2CO3=.100, NaOH=.02,HCO3=..42 fromquestion1
For Change:H2CO3=-.02, NaOH=-.02,HCO3=.+.02
For Equilium:H2CO3=.08, NaOH=0,HCO3=.40
2nd:Use Buffer equation: pH= -log(4.2x10-7)+log(.40÷.08)=?
3rd: answer:7pH
To determine the pH after adding 0.02 mol of NaOH to 1.0L of the buffer described in question #1, we need to calculate how much of the H2CO3 and NaHCO3 in the buffer will react with the added NaOH.
First, let's determine the number of moles of H2CO3 in the buffer. Given that the buffer has a concentration of 0.100 M and a volume of 1.0L, the number of moles of H2CO3 can be calculated using the formula:
moles = concentration × volume
moles of H2CO3 = 0.100 M × 1.0L
moles of H2CO3 = 0.100 mol
Since the ratio of H2CO3 to NaHCO3 in a buffer is 1:1, the number of moles of NaHCO3 in the buffer would also be 0.100 mol.
When NaOH is added to the buffer, it will react with the H2CO3 to form water and Na2CO3 according to the following balanced chemical equation:
H2CO3 + 2NaOH -> H2O + Na2CO3
Since the stoichiometric ratio is 1:2 between H2CO3 and NaOH, the 0.02 mol of NaOH will react with half of the H2CO3 in the buffer.
Therefore, the remaining moles of H2CO3 after the reaction will be:
remaining moles of H2CO3 = 0.100 mol - (0.02 mol / 2)
remaining moles of H2CO3 = 0.100 mol - 0.010 mol
remaining moles of H2CO3 = 0.090 mol
Similarly, the remaining moles of NaHCO3 after the reaction will be:
remaining moles of NaHCO3 = 0.100 mol - (0.02 mol / 2)
remaining moles of NaHCO3 = 0.100 mol - 0.010 mol
remaining moles of NaHCO3 = 0.090 mol
Now, let's calculate the moles of H2CO3 and NaHCO3 in the final volume of the buffer, which is still 1.0L.
moles of H2CO3 in final volume = moles of H2CO3 remaining = 0.090 mol
moles of NaHCO3 in final volume = moles of NaHCO3 remaining = 0.090 mol
Finally, to determine the new pH value, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
In this case, [A-] refers to the concentration of the conjugate base (NaHCO3) and [HA] refers to the concentration of the acid (H2CO3).
Given that the pKa value for H2CO3 is given as 4.2 x 10^-7, we can substitute the values into the equation:
pH = -log(4.2 x 10^-7) + log (0.090 mol / 0.090 mol)
pH = -(-6.38) + log (1)
pH = 6.38 + 0
pH = 6.38
Therefore, the pH after adding 0.02 mol of NaOH to 1.0L of the buffer described in question #1 will be 6.38.
To determine the pH when 0.02 mol of NaOH is added to 1.0L of the buffer described in question #1, we need to consider the reaction that occurs between NaOH and the components of the buffer (H2CO3 and NaHCO3).
First, let's calculate the moles of NaHCO3 required to prepare the buffer in question #1:
Step 1: Calculate the moles of H2CO3 in 1.0L of the buffer:
Given that the concentration of H2CO3 is 0.100 M and the volume of the buffer is 1.0L:
Moles of H2CO3 = concentration of H2CO3 x volume of the buffer
= 0.100 mol/L x 1.0L
= 0.100 mol
Step 2: Use the balanced chemical equation for the reaction between H2CO3 and NaHCO3 to determine the stoichiometric ratio:
H2CO3 + NaOH -> NaHCO3 + H2O
From the equation, we can see that 1 mol of H2CO3 reacts with 1 mol of NaHCO3. Therefore, the moles of NaHCO3 needed to react with the H2CO3 can be calculated as:
Moles of NaHCO3 = 0.100 mol
Step 3: Calculate the grams of NaHCO3 needed to prepare the buffer by using its molar mass:
The molar mass of NaHCO3 is calculated by summing the atomic masses of its constituent elements:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of NaHCO3 = (22.99 + 1.01 + 12.01 + (16.00 x 3)) g/mol
= 84.01 g/mol
Grams of NaHCO3 = Moles of NaHCO3 x Molar mass of NaHCO3
= 0.100 mol x 84.01 g/mol
= 8.401 g
Therefore, 8.401 grams of NaHCO3 should be added to one liter of 0.100 M H2CO3 to prepare a buffer with pH = 7.00.
Now, with this information and considering that 0.02 mol of NaOH is added to the 1.0L buffer, we can determine the pH.
Step 1: Calculate the moles of NaOH added:
Given that the moles of NaOH added is 0.02 mol.
Step 2: Calculate the moles of H2CO3 and NaHCO3 that react with the added NaOH:
Since the balanced chemical equation is 1:1 for H2CO3 and NaHCO3 reacting with NaOH:
Moles of H2CO3 reacted = Moles of NaHCO3 reacted = 0.02 mol
Step 3: Calculate the remaining moles of H2CO3 and NaHCO3:
Moles of H2CO3 remaining = Moles of H2CO3 initial - Moles of H2CO3 reacted
= 0.100 mol - 0.02 mol
= 0.080 mol
Moles of NaHCO3 remaining = Moles of NaHCO3 initial - Moles of NaHCO3 reacted
= 0.100 mol - 0.02 mol
= 0.080 mol
Step 4: Calculate the concentration of H2CO3 and NaHCO3 in the buffer after the reaction:
Concentration of H2CO3 = Moles of H2CO3 remaining / Volume of the buffer
= 0.080 mol / 1.0 L
= 0.080 M
Concentration of NaHCO3 = Moles of NaHCO3 remaining / Volume of the buffer
= 0.080 mol / 1.0 L
= 0.080 M
Step 5: Calculate the pH of the buffer:
To calculate the pH, we need to know the dissociation constant (Ka) for H2CO3. The given value is Ka = 4.2 x 10^(-7).
pH = pKa + log (concentration of NaHCO3 / concentration of H2CO3)
First, let's calculate pKa:
pKa = -log(Ka)
= -log(4.2 x 10^(-7))
≈ 6.38 (rounded to two decimal places)
Next, let's calculate the pH:
pH = 6.38 + log (0.080 M / 0.080 M)
= 6.38 + log(1)
= 6.38 + 0
= 6.38
Therefore, when 0.02 mol of NaOH is added to 1.0L of the buffer described in question #1, the pH of the resulting buffer is approximately 6.38.