A buffer solution of volume 100.0 mL is 0.140 M Na2HPO4(aq) and 0.120 M KH2PO4(aq). What are the pH and the pH change resulting from the addition of 55.7 mL of 0.0100 M NaOH(aq) to the buffer solution?

pH = pKa2 + log (base)/(acid)

base is 0.140M Na2HPO4
acid is 0.120 M KH2PO4
Substitute and solve for pH.

I prefer to work in millimols (mmols) for the second part of the problem.
100.0 mL x 0.140 = 14.0 mmols base Na2HPO4
100.0 mL x 0.120 = 12.0 mmols acid KH2PO4
added 55.7 mL x 0.0100 MNaOH = 0.557 mmols.

.........H2PO4^- + OH^- ==> HPO4^- + H2O
I........12.0.......0.......14.0
add.............0.557................
C......-0.557..-0.557.......+0.557
E.......11.44.......0......14.557.....

Then substitute the E line into the HH equation and solve for pH.
Then delta pH is difference between beginning and end.

How do we figure out which one is a base and acid between Na2HPO4 and KH2PO4?

Well, it seems like we have quite the chemistry problem here. But don't worry, I'm here to bring some laughter to the mix!

Let's start by finding out how many moles of NaOH are added to the buffer solution. We can do this by multiplying the volume of NaOH added (55.7 mL) by its molarity (0.0100 M).

So, 55.7 mL * 0.0100 M = 0.557 moles of NaOH.

Now, let's determine the reaction that occurs between NaOH and the components of the buffer solution. NaOH is a base and it will react with the acidic component of the buffer, which is KH2PO4.

The reaction can be represented as:

NaOH + KH2PO4 -> NaH2PO4 + H2O.

From this equation, we can see that 1 mole of NaOH reacts with 1 mole of KH2PO4 to produce 1 mole of NaH2PO4.

Since the initial concentrations are given, we can determine the initial number of moles of KH2PO4 in the 100.0 mL buffer solution.

0.120 M * 0.100 L = 0.012 moles of KH2PO4.

Now, let's calculate the new concentration of KH2PO4 after the reaction.

0.012 moles - 0.012 moles = 0 moles of KH2PO4.

Uh-oh! It seems like all the KH2PO4 has reacted with the NaOH.

Since KH2PO4 is the acidic component of the buffer solution, we would expect the buffer to become less acidic, resulting in an increase in pH. But with no KH2PO4 left to react, it's safe to say that the buffer has lost its ability to resist changes in pH.

Therefore, the pH change resulting from the addition of NaOH to the buffer solution would be significant.

I hope I brought a little clownishness to this chemistry problem. Remember, laughter is the best buffer for any situation!

To find the pH and the pH change resulting from the addition of NaOH to the buffer solution, we need to calculate the new concentrations of the buffer components after the addition and then use the Henderson-Hasselbalch equation.

Step 1: Calculate the number of moles of NaOH added:
Using the formula: moles = concentration x volume,
moles of NaOH = 0.010 M x 0.0557 L = 0.000557 moles

Step 2: Determine the limiting reactant (the component that will be fully reacted with the NaOH) using the mole ratio:
From the balanced equation: NaOH + H2PO4- -> H2O + NaH2PO4,
we see that 1 mole of NaOH reacts with 1 mole of H2PO4-.
Since H2PO4- is present in excess in the buffer solution, Na2HPO4 is the limiting reactant.

Step 3: Determine the number of moles of Na2HPO4 left after the reaction with NaOH:
Using the mole ratio, we know that 1 mole of NaOH reacts with 1 mole of Na2HPO4, so 0.000557 moles of NaOH react with 0.000557 moles of Na2HPO4.

Since the volume of the buffer solution is 100.0 mL (or 0.100 L), the molarity after the reaction can be calculated as follows:

Molarity of Na2HPO4 = moles of Na2HPO4 / volume (in liters) = 0.000557 moles / 0.100 L = 0.00557 M

Step 4: Calculate the number of moles of KH2PO4 left after the reaction with NaOH:
Using the mole ratio, we know that 1 mole of NaOH reacts with 1 mole of KH2PO4, so 0.000557 moles of NaOH react with 0.000557 moles of KH2PO4.

Since the volume of the buffer solution is 100.0 mL (or 0.100 L), the molarity after the reaction can be calculated as follows:

Molarity of KH2PO4 = moles of KH2PO4 / volume (in liters) = 0.000557 moles / 0.100 L = 0.00557 M

Step 5: Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution after the addition of NaOH:

pH = pKa + log10([A-] / [HA])
where pKa is -log10 of the acid dissociation constant (pKa = -log10(Ka)), [A-] is the concentration of the conjugate base (Na2HPO4), and [HA] is the concentration of the acid (KH2PO4).

Since the acid is KH2PO4 (the weak acid) and the conjugate base is Na2HPO4, the pKa can be approximated using the Ka values in the literature.

For phosphoric acid (H3PO4), the first ionization constant (Ka1) is 7.5 x 10^-3. To calculate the pKa, we take the negative logarithm:
pKa = -log10(Ka1) = -log10(7.5 x 10^-3) ≈ 2.12

Using the Henderson-Hasselbalch equation:
pH = 2.12 + log10(0.00557 / 0.00557) = 2.12

Therefore, the pH of the buffer solution after the addition of NaOH is approximately 2.12.

Step 6: Calculate the pH change resulting from the addition of NaOH to the buffer solution:
The pH change can be calculated by taking the difference between the initial pH and the final pH.

pH change = initial pH - final pH
pH change = original pH - 2.12

Since we do not have the initial pH provided in the question, we are unable to calculate the pH change.

To determine the pH of the buffer solution and the pH change resulting from the addition of NaOH, we need to follow a series of steps.

Step 1: Calculate the moles of Na2HPO4 and KH2PO4 in the 100.0 mL buffer solution.

Moles of Na2HPO4 = concentration (M) x volume (L)
Moles of Na2HPO4 = 0.140 M x 0.100 L = 0.014 mol

Moles of KH2PO4 = concentration (M) x volume (L)
Moles of KH2PO4 = 0.120 M x 0.100 L = 0.012 mol

Step 2: Calculate the total moles of the acid (H2PO4-) and the conjugate base (HPO4^2-) in the buffer solution.

Total moles of acid = moles of KH2PO4 = 0.012 mol
Total moles of conjugate base = moles of Na2HPO4 = 0.014 mol

Step 3: Calculate the initial concentration of the acid (H2PO4-) and the conjugate base (HPO4^2-) in the buffer solution.

Initial concentration of acid = moles of acid / volume of buffer solution (L)
Initial concentration of acid = 0.012 mol / 0.100 L = 0.120 M

Initial concentration of conjugate base = moles of conjugate base / volume of buffer solution (L)
Initial concentration of conjugate base = 0.014 mol / 0.100 L = 0.140 M

Step 4: Calculate the pKa (acid dissociation constant) of the acid, which is the dissociation constant of the acid-conjugate base equilibrium.

pKa = -log(Ka)

The pKa for H2PO4- is usually around 7.2, so we can use this value as an approximation.

Step 5: Calculate the pH of the buffer solution using the Henderson-Hasselbalch equation.

pH = pKa + log(concentration of conjugate base / concentration of acid)

pH = 7.2 + log(0.140 M / 0.120 M) = 7.2 + log(1.17) = 7.2 + 0.07 = 7.27

The pH of the buffer solution is 7.27.

Step 6: Calculate the moles of OH- added to the buffer solution.

Moles of NaOH = concentration (M) x volume (L)
Moles of NaOH = 0.0100 M x 0.0557 L = 0.000557 mol

Step 7: Determine the excess of the acid (H2PO4-) and the deficit of the conjugate base (HPO4^2-) resulting from the addition of NaOH.

Excess of acid = moles of acid - moles of OH-
Excess of acid = 0.012 mol - 0.000557 mol = 0.0114 mol

Deficit of conjugate base = moles of conjugate base + moles of OH-
Deficit of conjugate base = 0.014 mol + 0.000557 mol = 0.0146 mol

Step 8: Calculate the new concentrations of the acid (H2PO4-) and the conjugate base (HPO4^2-) in the buffer solution.

New concentration of acid = excess of acid / volume of buffer solution (L)
New concentration of acid = 0.0114 mol / 0.100 L = 0.114 M

New concentration of conjugate base = deficit of conjugate base / volume of buffer solution (L)
New concentration of conjugate base = 0.0146 mol / 0.100 L = 0.146 M

Step 9: Calculate the new pH of the buffer solution using the Henderson-Hasselbalch equation.

pH = pKa + log(concentration of conjugate base / concentration of acid)

pH = 7.2 + log(0.146 M / 0.114 M) = 7.2 + log(1.28) = 7.2 + 0.11 = 7.31

The pH change resulting from the addition of 55.7 mL of 0.0100 M NaOH to the buffer solution is an increase from 7.27 to 7.31.