1.0L of aqueous solution in which [H2CO3]=[HCO3^-]=0.10M and has [H^+]=4.2E-7. What is the concentration of [H^+] ofter 0.005 mole of NaOH has been added?
H2CO3 ==> H^+ + HCO3^-
k1 = (H^+)(HCO3^-)/(H2CO3)
I don't know if you are supposed to calculate or to look up k1. However, you can calculate it as follows.
Since (HCO3^-) = (H2CO3) [both are 0.1 M), then plugging in 0.l M for each gives k1 = (H^+).
If we start with 0.1 mol H2CO3 and add 0.005 mol NaOH, we have 0.1 - 0.005 mols H2CO3 remaining and an extra 0.005 mol HCO3^- formed to make the final (HCO3^-)= 0.1 + 0.005 = ??
Plug k1, H2CO3, and HCO3^- into the k1 expression above and solve for (H^+). I found 3.97 x 10^-7. Post your work if you have trouble. By the way, it is a longwe way but you can also use the Henderson-Hasselbalch equation. Note also that this is a buffer solution and the H^+ didn't change much even though an amount of NaOH equivalent to 5% was added. That's what it is supposed to do. 0.005 mol NaOH added to an unbuffered solution would change the H^+ to 2 x 10^-12.
After the addition of NaOH to the solution, the concentration of H2CO3 is reduced to 0.095 M (0.1 - 0.005) and the concentration of HCO3^- is increased to 0.105 M (0.1 + 0.005).
We can use the k1 expression from before: k1 = (H^+)(HCO3^-)/(H2CO3)
Now we have k1 = (H^+)(0.105)/(0.095)
We already calculated k1 to be 4.2E-7, so now we can solve for the new [H^+]:
4.2E-7 = (H^+)(0.105)/(0.095)
(H^+) = 4.2E-7 * (0.095/0.105)
(H^+) = 3.8E-7
The new concentration of [H^+] after the addition of NaOH is 3.8E-7 M.
To calculate the concentration of [H^+] after 0.005 moles of NaOH has been added, we need to consider the stoichiometry of the reaction between NaOH and H2CO3.
The balanced equation for the reaction is:
2HCO3^- + 2Na^+ + 2H2O -> 2CO3^2- + 2Na^+ + 2H2O
From the equation, we can see that 2 moles of HCO3^- react with 1 mole of NaOH. Therefore, if 0.005 moles of NaOH is added, it will react with 0.01 moles of HCO3^-.
Initially, we have [HCO3^-] = 0.10 M. After the reaction, the final concentration of HCO3^- will be:
[HCO3^-] = (initial moles - moles reacted) / final volume
The final volume is the initial volume plus the volume of NaOH added. But since the volume is not given, we will assume that it remains the same. Therefore, the final volume is still 1.0 L.
[HCO3^-] = (0.10 M - 0.01 mol) / 1.0 L
[HCO3^-] = 0.09 M
Since we are assuming that [HCO3^-] = [H2CO3], the concentration of H2CO3 is also 0.09 M.
Now we can use the expression for k1 to calculate [H^+]:
k1 = (H^+)(HCO3^-) / (H2CO3)
Rearranging the equation, we get:
(H^+) = k1 * (H2CO3) / (HCO3^-)
Plugging in the known values, we have:
(H^+) = (3.97 x 10^-7) * (0.09 M) / (0.09 M)
(H^+) = 3.97 x 10^-7
Therefore, the concentration of [H^+] after 0.005 moles of NaOH has been added is 3.97 x 10^-7 M.
To calculate the concentration of [H^+] after 0.005 moles of NaOH has been added, we can follow the steps provided in your explanation.
Step 1: Calculate the initial value of (HCO3^-) after adding NaOH.
Since [HCO3^-] = [H2CO3] = 0.10 M, adding 0.005 moles of NaOH will result in an increase in (HCO3^-) by 0.005 moles.
Therefore, the final value of (HCO3^-) is 0.10 + 0.005 = 0.105 M.
Step 2: Calculate the value of k1 using the expression:
k1 = (H^+)(HCO3^-)/(H2CO3)
Since [HCO3^-] = (H2CO3) = 0.10 M, we can substitute these values into the expression:
k1 = (H^+)(0.10)/(0.10)
Simplifying, k1 = (H^+).
Step 3: Substitute the values of k1, (HCO3^-), and (H2CO3) into the k1 expression and solve for (H^+).
0.10 = (H^+)
Therefore, the initial concentration of [H^+] is 0.10 M.
Step 4: Calculate the final concentration of [H^+] after adding NaOH.
Since a buffer solution resists large changes in pH, the concentration of [H^+] is expected to remain relatively constant.
Therefore, the final concentration of [H^+] is still approximately 0.10 M after adding 0.005 moles of NaOH.
Please note that these calculations assume that the volume of the solution remained constant throughout the process.