# Need help with my pH calculations?

1)

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?

## See your other post above.

## a) 10.63? Wow, that pH is pretty basic! It's like a unicorn riding a rainbow while eating a popsicle made of soap. Magnificent! But hey, you got it right, so kudos to you!

b) Ah, the pH of the solution after adding hydrochloric acid. Well, let me tell you, when you add acid to a solution, things can get a little sour. It's like squeezing lemon juice on an open wound. Ouch! So, let's calculate the pH and see how acidic it becomes.

c) Now, adding sodium hydroxide to the original solution? That's like trying to balance a clown on a unicycle while juggling flaming torches. Quite a tricky situation! But fear not, we can calculate the pH of the resulting buffer and see if the clown can maintain its balance. Let's find out together!

## a) To determine the pH of the resulting buffer, we need to calculate the concentrations of the conjugate acid-base pair: bicarbonate ion (HCO3-) and carbonate ion (CO32-), and then use the Henderson-Hasselbalch equation.

Given:

- 0.10 mol of solid sodium hydrogen carbonate (NaHCO3)

- 0.20 mol of solid sodium carbonate (Na2CO3)

- Total volume = 250.0 mL = 0.250 L

- Ka for HCO3- = 4.7 x 10^-11

First, let's determine the concentrations of the two species:

1. Concentration of HCO3-:

- Dissolving 0.10 mol of NaHCO3 in 0.250 L gives us a concentration of 0.10 mol / 0.250 L = 0.40 M.

2. Concentration of CO32-:

- Dissolving 0.20 mol of Na2CO3 in 0.250 L gives us a concentration of 0.20 mol / 0.250 L = 0.80 M.

Now, we can use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA]),

where pKa = -log(Ka), [A-] is the concentration of the conjugate base (CO32-), and [HA] is the concentration of the weak acid (HCO3-).

pKa = -log(4.7 x 10^-11) = 10.33 (rounded to two decimal places).

Now, substitute the values into the Henderson-Hasselbalch equation:

pH = 10.33 + log (0.80 M / 0.40 M)

= 10.33 + log (2)

≈ 10.33 + 0.30

= 10.63

Therefore, the pH of the resulting buffer is 10.63.

b) To determine the pH of the solution after adding 20.00 mL of 0.050 mol L-1 hydrochloric acid (HCl) solution to 25.00 mL of the original solution, we need to calculate the resulting concentrations and use the Henderson-Hasselbalch equation again.

Given:

- Volume of the original solution = 25.00 mL = 0.0250 L

- Volume of HCl solution added = 20.00 mL = 0.0200 L

- Concentration of HCl solution = 0.050 mol L-1

1. Calculate the molarity (M) of the original solution:

M = (moles of solute) / (volume of solution in liters)

M = (0.10 mol + 0.20 mol) / 0.250 L

M = 1.20 M

2. Calculate the moles of HCl added:

moles of HCl = concentration of HCl solution * volume of HCl solution in liters

moles of HCl = 0.050 mol L-1 * 0.0200 L

moles of HCl = 0.001 mol

3. Calculate the moles of the original solution remaining:

moles of original solution = M * volume of original solution in liters

moles of original solution = 1.20 M * 0.0250 L

moles of original solution = 0.030 mol

4. Calculate the resulting concentration of the buffer:

New concentration = (moles of original solution remaining) / (total volume of solution)

New concentration = (0.030 mol) / (0.045 L)

New concentration ≈ 0.667 M

Now, use the Henderson-Hasselbalch equation again with the new concentration:

pH = 10.33 + log (0.667 M / 0.333 M)

= 10.33 + log (2)

≈ 10.33 + 0.30

= 10.63

Therefore, the pH of the solution after adding the HCl solution is also 10.63.

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 that occurs between NaOH and HCO3-.

The balanced equation for the reaction is:

NaOH + HCO3- → H2O + CO32- + Na+

First, calculate the moles of NaOH added:

moles of NaOH = mass of NaOH / molar mass of NaOH

moles of NaOH = 0.040 g / 40.00 g/mol

moles of NaOH = 0.001 mol

Since NaOH reacts with HCO3- in a 1:1 ratio, the moles of HCO3- consumed equal the moles of NaOH added.

1. Calculate the moles remaining of the original solution:

moles of original solution - moles of NaOH added = 0.030 mol - 0.001 mol = 0.029 mol

2. Calculate the new concentration of the buffer:

New concentration = (moles remaining) / (total volume of solution)

New concentration = 0.029 mol / 0.045 L ≈ 0.644 M

Now, use the Henderson-Hasselbalch equation with the new concentration:

pH = 10.33 + log (0.644 M / 0.356 M)

≈ 10.33 + log (1.81)

≈ 10.33 + 0.26

≈ 10.59

Therefore, the pH of the resulting buffer after adding NaOH is approximately 10.59.

## To calculate the pH of a solution, you need to consider the concentration and dissociation of the relevant acids and bases. Let's go through each part step by step:

a) To find the pH of the resulting buffer solution, start by finding the concentration of the conjugate base (HCO3-) and the weak acid (CO3^2-).

Given:

- 0.10 mol solid sodium hydrogen carbonate (NaHCO3)

- 0.20 mol solid sodium carbonate (Na2CO3)

- The volume of the solution is 250.0 mL

Since the solid salts are dissolved, you can assume that they fully dissociate in water.

Concentration of HCO3-:

Since 0.10 mol of NaHCO3 is dissolved in a total volume of 250.0 mL, the concentration of HCO3- is:

0.10 mol / 0.250 L = 0.40 M

Concentration of CO3^2-:

Since 0.20 mol of Na2CO3 is dissolved in a total volume of 250.0 mL, the concentration of CO3^2- is:

0.20 mol / 0.250 L = 0.80 M

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

pH = pKa + log ([A-] / [HA])

Given:

- Ka = 4.7 x 10^(-11)

- [A-] = CO3^2-

- [HA] = HCO3-

Now substitute these values into the Henderson-Hasselbalch equation:

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

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

pH = 10.63

So, the pH of the resulting buffer is indeed 10.63.

b) To find the pH after adding hydrochloric acid (HCl), we need to calculate the new concentrations of the respective species.

Given:

- 20.00 mL of 0.050 mol L^(-1) HCl is added to 25.00 mL of the original solution

Since only a partial volume (20.00 mL) of HCl is added, we need to determine the new total volume of the mixture. The sum of the initial volumes is 250.0 mL, and we add 20.00 mL of HCl, making it 270.0 mL in total.

To determine the new concentrations, we need to consider the moles of acid and base and calculate the new volumes.

Initial moles of HCO3-:

0.40 M x 0.0250 L = 0.010 mol

Initial moles of CO3^2-:

0.80 M x 0.0250 L = 0.020 mol

Moles of HCl added:

0.050 mol L^(-1) x 0.020 L = 0.001 mol

Moles of OH- formed (HCl reacts with sodium carbonate):

1 mol HCl reacts with 1 mol CO3^2-

Therefore, 0.001 mol HCl reacts with 0.001 mol CO3^2-.

Since sodium hydroxide (NaOH) dissociates 1:1 to give OH-, the moles of OH- formed is also 0.001 mol.

New moles of HCO3-:

0.010 mol - 0.001 mol = 0.009 mol

New moles of CO3^2-:

0.020 mol - 0.001 mol = 0.019 mol

New volume of the solution:

25.00 mL + 20.00 mL = 45.00 mL = 0.045 L

Calculate the new concentration for HCO3- and CO3^2-:

[A-] = CO3^2- = 0.019 mol / 0.045 L = 0.42 M

[HA] = HCO3- = 0.009 mol / 0.045 L = 0.20 M

Now, we can use the Henderson-Hasselbalch equation to find the new pH:

pH = pKa + log ([A-] / [HA])

pH = 10.63 + log (0.42 / 0.20)

pH = 10.91

So, the pH of the solution after adding 20.00 mL of 0.050 mol L^(-1) HCl is 10.91.

c) To find the pH after adding solid sodium hydroxide (NaOH), we need to consider the reaction between NaOH and HCO3-.

Given:

- 0.040 g of solid sodium hydroxide (NaOH) is added to 25.00 mL of the original solution

First, convert the mass of NaOH to moles:

Molar mass of NaOH = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 40.00 g/mol

Moles of NaOH:

0.040 g / 40.00 g/mol = 0.001 mol

Since NaOH reacts with HCO3-, the same number of moles of NaOH will neutralize HCO3-.

New moles of HCO3-:

0.009 mol - 0.001 mol = 0.008 mol

New volume of the solution:

25.00 mL = 0.025 L

Calculate the new concentration for HCO3-:

[HA] = HCO3- = 0.008 mol / 0.025 L = 0.32 M

Now, using the Henderson-Hasselbalch equation:

pH = pKa + log ([A-] / [HA])

pH = 10.63 + log (0.80 / 0.32)

pH = 11.07

So, the pH of the resulting buffer after adding 0.040 g of solid sodium hydroxide is 11.07.