If 10 mL of 1M NaOH are added to one liter of a buffer that is 0.3 M acetic acid and 0.2 M sodium acetate (Na+CH3COO-), how much does the pH change? The pKa for acetic acid is 4.76.
To determine how much the pH changes after adding NaOH to the buffer solution, we need to calculate the new concentrations of acetic acid and acetate ion.
First, let's calculate the initial concentrations of acetic acid (CH3COOH) and acetate ion (CH3COO-) in the buffer before adding NaOH.
Acetic acid (CH3COOH) concentration = 0.3 M
Acetate ion (CH3COO-) concentration = 0.2 M
Next, we need to determine the change in concentration of acetic acid and acetate ion after adding NaOH.
NaOH is a strong base and reacts with acetic acid (a weak acid) to form sodium acetate and water.
The balanced chemical equation for this reaction is:
CH3COOH + OH- -> CH3COO- + H2O
Since NaOH is 1 M, 10 mL of NaOH is equivalent to 0.01 moles.
The stoichiometry of the reaction shows that 1 mole of NaOH reacts with 1 mole of CH3COOH to produce 1 mole of CH3COO-.
Therefore, after adding NaOH, the concentration of acetate ion (CH3COO-) will increase by 0.01 M.
Now, let's calculate the new concentrations of acetic acid and acetate ion.
Acetic acid (CH3COOH) concentration = 0.3 M - 0.01 M = 0.29 M
Acetate ion (CH3COO-) concentration = 0.2 M + 0.01 M = 0.21 M
To find out how much the pH changes, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([CH3COO-] / [CH3COOH])
The pKa for acetic acid is given as 4.76.
Using the Henderson-Hasselbalch equation, plug in the values:
pH = 4.76 + log (0.21 / 0.29)
Calculating the logarithm:
pH = 4.76 + log (0.724)
Using a calculator, take the logarithm:
pH = 4.76 - 0.1434
Finally, calculate the pH:
pH = 4.6166
The pH changes from its initial value to approximately 4.62 after adding 10 mL of 1 M NaOH to the buffer solution.
To find out how much the pH changes when 10 mL of 1M NaOH is added to the buffer, we need to consider the reaction that takes place between the NaOH and the acetic acid in the buffer.
The reaction between NaOH and acetic acid can be represented as follows:
NaOH + CH3COOH -> CH3COONa + H2O
In this reaction, the NaOH reacts with acetic acid to form sodium acetate (CH3COONa) and water. This reaction is a neutralization reaction.
To determine the change in pH, we need to calculate the change in concentration of acetic acid (CH3COOH) and sodium acetate (CH3COONa).
Given that the initial concentration of acetic acid is 0.3 M and sodium acetate is 0.2 M, and that 10 mL of 1M NaOH is added, we can calculate the amount of acetic acid and sodium acetate that reacts with the NaOH.
Since NaOH and acetic acid react in a 1:1 mole ratio, 10 mL (0.01 L) of 1M NaOH reacts with 0.01 moles of acetic acid.
Now, we can calculate the final concentrations of acetic acid and sodium acetate. The initial concentration of acetic acid is 0.3 M, and 0.01 moles reacted, so the final concentration of acetic acid is:
Final concentration of acetic acid = (0.3 M - 0.01 moles) / 1 L
Similarly, the final concentration of sodium acetate can be calculated by considering the initial concentration and the amount of sodium acetate formed:
Final concentration of sodium acetate = (0.2 M + 0.01 moles) / 1 L
Now, we can use the Henderson-Hasselbalch equation to calculate the new pH value:
pH = pKa + log ([A-] / [HA])
Where pKa is the negative logarithm of the acid dissociation constant (pKa = -log[Ka]), [A-] is the concentration of the conjugate base (sodium acetate), and [HA] is the concentration of the acid form (acetic acid).
Using the given pKa of 4.76 and the calculated concentrations of sodium acetate and acetic acid, we can plug them into the Henderson-Hasselbalch equation to find the new pH value.
pH = 4.76 + log ([0.2 M + 0.01 moles] / [0.3 M - 0.01 moles])
By calculating the above equation, we can determine the new pH value.
Please note that this calculation assumes that the volume change resulting from the addition of NaOH is negligible compared to the original volume of the buffer solution.