# Not sure if i posted this but

Need help with my pH calculations?

0.10 mol of solid sodium hydrogen carbonate and 0.20 mol of solid sodium carbonate are
dissolved in the same beaker of water, transferred to a volumetric ﬂask and made to 250.0
mL. The Ka for HCO3
– is 4.7 x 10–11.

a) What is the pH of the resulting buﬀer?
i got pH i calculated here is 10.63

b) What is the pH of solution after 20.00 mL of 0.050 mol L–1 hydrochloric acid solution is
added to 25.00 mL of the original solution?

c)What is the pH of the resulting buﬀer after 0.040 g of solid sodium hydroxide is added
to 25.00 mL of the original solution?

thx

## a is ok.

b. You have how much in solution.
100 millimols Na2CO3 x (25 mL/250 mL) = 10
200 millimols NaHCO3 x (25/250) = 20
You are adding 20 mL x 0.050M = 1 millimol HCl.

.........CO3^2- + H^+ ==> HCO3^-
I........10........0.......20
C........-1.......-1.......+1
E.........9........0........21
Substitute the E line into the HH equation and solve for pH.

c is done the same way.
........HCO3 + OH^= ==> CO3^2- + H2O
etc
etc

## To calculate the pH of the resulting buffer, we need to consider the dissociation of sodium hydrogen carbonate (NaHCO3) and the equilibrium reaction between bicarbonate ion (HCO3-) and water (H2O):

NaHCO3(s) -> Na+(aq) + HCO3-(aq)

HCO3-(aq) + H2O(l) ⇌ H2CO3(aq) + OH-(aq)

a) To determine the initial concentrations of HCO3- and CO3^2-, we use the molarities of NaHCO3 and Na2CO3 in the mixture:

NaHCO3:
Moles of NaHCO3 = 0.10 mol (given)
Initial volume = 250.0 mL = 0.250 L

Initial concentration of NaHCO3 = (0.10 mol) / (0.250 L) = 0.40 mol/L (M)

Na2CO3:
Moles of Na2CO3 = 0.20 mol (given)
Initial volume = 250.0 mL = 0.250 L

Initial concentration of Na2CO3 = (0.20 mol) / (0.250 L) = 0.80 mol/L (M)

Next, we need to calculate the concentrations of HCO3- and CO3^2- ions after the dissociation of NaHCO3 and partially-buffernized carbonate ions:

HCO3-:
Since NaHCO3 is a strong electrolyte, it completely dissociates in water to form Na+ and HCO3- ions. Therefore, the initial concentration of HCO3- is equal to the initial concentration of NaHCO3:

[HCO3-] = 0.40 M

CO3^2-:
CO3^2- ions are the partially-buffed form of carbonate ions, so we need to consider the reaction between HCO3- and water. From the balanced equation, we can see that the initial concentration of CO3^2- is equal to the initial concentration of Na2CO3:

[CO3^2-] = 0.80 M

Now, we can calculate the pH of the resulting buffer using the Henderson-Hasselbalch equation:

pH = pKa + log([CO3^2-] / [HCO3-])

The pKa given is 4.7 x 10^(-11).

Substituting the values into the equation:

pH = -log(4.7 x 10^(-11)) + log(0.80 / 0.40)

Calculating this expression gives a pH of approximately 10.63.

b) To determine the pH after adding 20.00 mL of 0.050 mol/L hydrochloric acid (HCl) solution to 25.00 mL of the original solution, we need to consider the reaction between HCl and the NaHCO3/Na2CO3 buffer solution.

First, calculate the moles of HCl added:
Moles of HCl = (0.050 mol/L) x (0.020 L) = 0.001 mol

Next, determine the new volumes of the buffer solution:
Initial volume = 25.00 mL = 0.025 L
Added volume of HCl = 20.00 mL = 0.020 L

Total volume after addition = Initial volume + Added volume = 0.025 L + 0.020 L = 0.045 L

Now, calculate the new concentrations of HCO3- and CO3^2- ions:

[HCO3-] = (0.40 mol/L x 0.025 L) / 0.045 L = 0.222 M
[CO3^2-] = (0.80 mol/L x 0.025 L) / 0.045 L = 0.444 M

Substituting these values into the Henderson-Hasselbalch equation:

pH = pKa + log([CO3^2-] / [HCO3-])

Using the given pKa (4.7 x 10^(-11)), we can calculate the new pH.

c) To determine the pH of the resulting buffer after adding 0.040 g of solid sodium hydroxide (NaOH) to 25.00 mL of the original solution, we need to consider the reaction between NaOH and the NaHCO3/Na2CO3 buffer solution.

First, calculate the moles of NaOH added:
Moles of NaOH = (0.040 g) / (molar mass of NaOH)

Next, determine the new volumes of the buffer solution:
Initial volume = 25.00 mL = 0.025 L

Total volume after addition = Initial volume = 0.025 L

Now, calculate the new concentrations of HCO3- and CO3^2- ions:

[HCO3-] = (0.40 mol/L x 0.025 L) / 0.025 L = 0.40 M
[CO3^2-] = (0.80 mol/L x 0.025 L) / 0.025 L = 0.80 M

Substituting these values into the Henderson-Hasselbalch equation:

pH = pKa + log([CO3^2-] / [HCO3-])

Using the given pKa (4.7 x 10^(-11)), we can calculate the new pH.

## To solve these pH calculations, we need to apply the principles of acid-base chemistry and buffer solutions. Here's how you can approach each part of the question:

a) Finding the pH of the resulting buffer:
A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) and helps resist changes in pH. In this case, the weak acid is HCO3- and its conjugate base is CO3^2-.

To find the pH of a buffer solution, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])

In this equation, pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

Given the Ka for HCO3- is 4.7 x 10^-11, you can use this value to calculate the pKa (-log(Ka)).

Using the concentrations provided (0.10 mol of HCO3- and 0.20 mol of CO3^2-) and the formula volume (250.0 mL), convert the given concentrations to molarities (M).

Substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([CO3^2-]/[HCO3-])

By calculating the pH using the given values, you should obtain a pH of 10.63 as you mentioned.

b) Finding the pH after adding hydrochloric acid solution:
To find the new pH after adding hydrochloric acid solution, we need to consider the reaction between the strong acid (HCl) and the weak acid (HCO3-) in the buffer solution. When a strong acid is added to a buffer, it reacts with the weak acid, consuming it and leading to an increased concentration of the conjugate base.

To solve this, we can consider the equilibrium reaction between HCO3- and HCl:
HCO3- + HCl → H2CO3 + Cl-

Since the concentration of HCl is given as 0.050 mol L^-1 and we are adding 20.00 mL, we can calculate the moles of HCl added. Similarly, we can calculate the moles of HCO3- initially present in the 25.00 mL solution.

Taking the sum of the moles of HCO3- and the moles of Cl- after the reaction, calculate the new concentrations of HCO3- and CO3^2- in the buffer solution by dividing the moles by the final volume of the solution (45.00 mL).

Finally, use the Henderson-Hasselbalch equation with the new concentrations of [CO3^2-] and [HCO3-] to calculate the new pH.

c) Finding the pH after adding sodium hydroxide:
To find the new pH after adding solid sodium hydroxide (NaOH), we need to consider the reaction between the strong base (NaOH) and the acidic components of the buffer solution (HCO3- and H2CO3).

To solve this, we can consider the reaction between NaOH and HCO3-:
NaOH + HCO3- → NaHCO3 + OH-

Since the solid sodium hydroxide is given as 0.040 g, convert this to moles of NaOH. Then, calculate the moles of NaOH reacting with HCO3- based on the stoichiometry of the reaction.

Calculate the new concentrations of HCO3-, CO3^2-, and OH- in the buffer solution using the calculated moles and the new final volume (25.00 mL + 0.040 g NaOH).

Finally, use the Henderson-Hasselbalch equation with the new concentrations of [CO3^2-] and [HCO3-] to calculate the new pH.

By following these steps for parts b) and c), you should be able to determine the pH of the resulting solution in each case.