1L of a buffer composed of NaHC2O4 and Na2C2O4 has a pHof 4.0 and a total molarity of 1.
a). what would the effect on the pH be of adding 1 gram of sodium hydrogen oxalate to this buffer?
b). what would be the effect on the pH of adding 1 gram of sodium oxalate to the original buffer?
c). what would be the effect on the pH of adding 10mL of 2M sodium hydroxide to the original buffer?
d). what would be the effect on the pH of adding 100mL of 2M sodium hydroxide to the original buffer?
Ashley, tel me what your problem is so I will know how to answer these.
But to get you stated, if you add an acid the soln will become more acid. If you add a base the soln will become more basic.
When you say "effect" is it enough to say it becomes more acid or more basic or are you to calculate the pH at the new conditions. If the latter, I can show you how to do one of them and the others follow.
i am suppose to calculate the pH at the new conditions, but i am super lost. also for c and d i think you would have to figure out the moles, but then i do not know what to do from there.
i am suppose to calculate the pH at the new conditions, but i am super lost. also for c and d i think you would have to figure out the moles, but then i do not know what to do from there.
Thanks.
You have 1L of a 1 M solution of NaHC2O4 and Na2C2O4 a a pH = 4.0
a. You add 1g NaHC2O4 to this.
First you must calculate what you have in the buffer. You do this as follows:
base(B) + acid(A) = 1M
That's two unknowns so you need a second equation. That one is
pH = pK2 (for H2C2O4) + log (B)/(A)
4.0 = 4.27 + log(B)/(A)
Look up pK2 for H2C2O4 and use that value from your notes/text and not the one above. Texts differ and mine is several years old.
There are your two equations in two unknowns. Solve for A and B.
Then add the 1 g (converted to mols and M of course) NaHC2O4
Here is what I get.
4.0 = 4.27 + log b/a
b/a = 0.513 pr b = 0.513a
Then
a + b = 1
a + 0.513a = 1
1.513a = 1
a = 0.661 M
b = 1-0.661 = 0.339 M
Therefore, the original solution consists of 1L of 0.661M NaHC2O4 and 0.339M Na2C2O4. You may obtain slightly different numbers if you use a different pKa2 than I used above. Check my work. The a part of the problem follows:
1g NaHC2O4 = 1/molar mass = approximately 0.00893 in 1L which makes 0.00893M.(It appears I added molarity; actually I multiplied 1L x M, added moles, and divided by 1L which is the same as adding molarities). We had 0.661M NaHC2O4. Total mols/L = 0.661+0.00893 = 0.670
(It may appear I added molarity; actually I multiplied 1L x M, added moles, and divided by 1L which is the same as adding molarities).
pH = 4.27 + log(0.339/0.670) =3.99
Remember to redo this with your own pKa2 and check my work too.
The b part is done the same way but will be slightly more basic. The c and d parts are best worked with an ICE chart and use mols. If you need further help please explain exactly what you don't understand and show work so I can see where you're headed.
for a i got pH=3.97 and for b i got 3.99
for c and d i am stuck.
To answer these questions, we need to understand the concept of buffer solutions and how they work.
A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid. It helps to maintain the pH of a solution by resisting changes in pH when small amounts of acidic or basic substances are added.
Let's go through each question one by one:
a) What would be the effect on the pH of adding 1 gram of sodium hydrogen oxalate to this buffer?
Sodium hydrogen oxalate (NaHC2O4) is an acidic substance. When added to the buffer, it will react with the basic component of the buffer (Na2C2O4) to form additional acid. This will increase the concentration of the acid component in the buffer.
As a result, the buffer will have a higher concentration of the acid compared to its base, shifting the equilibrium towards the acidic side. This will cause a decrease in pH, making the solution more acidic.
b) What would be the effect on the pH of adding 1 gram of sodium oxalate to the original buffer?
Sodium oxalate (Na2C2O4) is a basic substance. When added to the buffer, it will react with the acidic component of the buffer (NaHC2O4) to form additional base. This will increase the concentration of the base component in the buffer.
As a result, the buffer will have a higher concentration of the base compared to its acid, shifting the equilibrium towards the basic side. This will cause an increase in pH, making the solution more basic.
c) What would be the effect on the pH of adding 10 mL of 2M sodium hydroxide to the original buffer?
Sodium hydroxide (NaOH) is a strong base. When added to the buffer, it will react with the weak acid component of the buffer (NaHC2O4) to form water and the salt NaC2O4.
The reaction will consume some of the weak acid, reducing its concentration. However, the buffer will still have a significant concentration of its acidic component, allowing it to resist changes in pH.
The addition of sodium hydroxide will increase the concentration of the conjugate base (NaC2O4) in the buffer, shifting the equilibrium towards the basic side. This will cause a slight increase in pH, making the solution slightly more basic.
d) What would be the effect on the pH of adding 100 mL of 2M sodium hydroxide to the original buffer?
Similar to the previous scenario, the addition of sodium hydroxide will react with the weak acid component (NaHC2O4) to form water and the salt NaC2O4. The buffer will still have some concentration of the weak acid, but it will be significantly reduced.
This time, the increase in the concentration of the conjugate base (NaC2O4) will be more substantial than before. As a result, the equilibrium will shift further towards the basic side, causing a greater increase in pH compared to adding only 10 mL of sodium hydroxide.
Please note that the exact change in pH will depend on the specific pKa of the acid component and the initial concentrations of the buffer components. These calculations can be done using the Henderson-Hasselbalch equation, which relates the pH of a buffer to the concentrations of its components.