how many mL of .100 M of acetic acid and .100 M of NaOH, would be need to make the volume 1.0 liters and have a pH7.00 if Ka=1.8x10^-5
please write out the eq for me
To determine how many milliliters of 0.100 M acetic acid (CH3COOH) and 0.100 M sodium hydroxide (NaOH) are needed to make a 1.0 liter solution with a pH of 7.00, we need to understand the equations involved.
The first equation is the dissociation of acetic acid in water:
CH3COOH ⇌ H+ + CH3COO-
The second equation is the dissociation of water into H+ and OH- ions:
H2O ⇌ H+ + OH-
The third equation is the reaction between sodium hydroxide and acetic acid:
CH3COOH + NaOH ⇌ H2O + CH3COONa
Now, we need the Ka value (acid dissociation constant) for acetic acid, which is 1.8x10^-5.
Using the given information, we can calculate the concentrations of H+ and OH- ions in the solution and determine the pH.
Now, let's begin by calculating the concentration (in moles per liter) of H+ and OH- ions in a solution of pure water at 25°C. Given that the concentration of water, H2O, is essentially constant at 55.5 M:
[H2O] = 55.5 M
[H+] = [OH-] = x
As the water molecules dissociate into equal amounts of H+ and OH- ions:
[H2O] - [H+] = [OH-]
55.5 - x = x
2x = 55.5
x = 27.75 M
So, the concentration of H+ ions (and also OH- ions) in pure water is 27.75 M.
To make a solution with a pH of 7.00, which is neutral, the concentration of H+ and OH- ions needs to be equal. Therefore, we want a concentration of 27.75 M for both H+ and OH- ions.
Now, since acetic acid is a weak acid, only a fraction of it dissociates. We need to calculate the concentration of acetic acid, CH3COOH (in moles per liter), given that its dissociation constant, Ka, is 1.8x10^-5.
The equation for the dissociation of acetic acid can be written as:
CH3COOH ⇌ H+ + CH3COO-
Let's assume that the concentration of CH3COOH that dissociates is x (moles per liter). Therefore, the concentration of H+ ions formed will also be x, while the concentration of undissociated acetic acid will be the initial concentration minus x.
[CH3COOH] - x = initial concentration - x
[H+] = x
Given that the initial concentration of acetic acid is 0.100 M, we can write the equation:
0.100 - x = x
2x = 0.100
x = 0.050 M
Therefore, the concentration of acetic acid that dissociates, and consequently the concentration of H+ ions produced, is 0.050 M.
Now, to achieve a pH of 7.00, which corresponds to a concentration of 27.75 M for both H+ and OH- ions, we need to determine the amount of NaOH required to neutralize the H+ ions and provide an equal concentration of OH- ions.
Since NaOH is a strong base that fully dissociates, the concentration of OH- ions produced will be equal to the concentration of NaOH used.
Let's assume the volume (in liters) of NaOH required is V liters, and the concentration of NaOH is 0.100 M.
[NaOH] = [OH-] = V
Using the stoichiometry of the reaction, we know that the ratio of acetic acid (CH3COOH) to NaOH is 1:1. Therefore, the moles of NaOH used will be equal to the moles of acetic acid that dissociates.
Since the volume of the final solution is 1.0 L, the moles of NaOH used can be calculated as:
moles of NaOH = V x [NaOH]
moles of NaOH = V x 0.100
moles of acetic acid that dissociates = moles of NaOH used
moles of acetic acid that dissociates = 0.050
Therefore, we can set up the equation:
0.050 = V x 0.100
Solving for V:
V = 0.050 / 0.100
V = 0.500 L
To convert the volume to milliliters, we multiply by 1000:
V = 0.500 x 1000 mL
V = 500 mL
So, to make a 1.0 liter solution with a pH of 7.00, you will need 500 mL of 0.100 M NaOH and 500 mL of 0.100 M acetic acid.
Remember to always check your calculations and pay attention to significant figures in your final answer.