How many milliliters of 1.0 M NaOH must be added to 200. mL of 0.10 M NaH2PO4 to make

a buffer solution with a pH of 7.50? (5 points)

Why did the chemist open a bakery? Because he kneaded a new career and wanted to rise to the occasion!

To calculate the amount of NaOH needed, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of the acid and base components.

First, we need to find the ratio of the concentrations of NaH2PO4 and Na2HPO4 in the buffer solution. NaH2PO4 is a weak acid that will donate one proton (H+) to become HPO4^2-, while NaOH is a strong base that will accept that proton to form water.

Since the pH of 7.50 is slightly basic, we want to have a slightly greater concentration of the weak acid H2PO4^- in the solution, compared to the conjugate base HPO4^2-.

Now, let's do the math (albeit the fun kind):

Given:
Volume of NaH2PO4 solution (Va) = 200 mL
Concentration of NaH2PO4 (Ca) = 0.10 M
Concentration of NaOH (Cb) = 1.0 M
Final pH of the buffer solution (pHf) = 7.50

Henderson-Hasselbalch equation:
pH = pKa + log([base]/[acid])

The pKa of the phosphate buffer system is 7.21 (you can check if you want).

To convert from mL to L, we divide by 1000.

Now, let the volume of NaOH solution needed be Vb mL.

So, the volume of NaOH solution needed is:
Vb = (Ca*Va*(pHf - pKa))/(Cb - Ca)

Plug in the given values:
Vb = (0.10 M * (200 mL / 1000 mL) * (7.50 - 7.21)) / (1.0 M - 0.10 M)

Calculating... *drumroll*
Vb ≈ 0.109 mL

So, you would need approximately 0.109 mL of 1.0 M NaOH to add to your 200 mL of 0.10 M NaH2PO4 to create a buffer solution with a pH of 7.50.

Remember to do it with caution, a little bit goes a long way!

To calculate the amount of NaOH needed to make a buffer solution with a pH of 7.50, we need to use the Henderson-Hasselbalch equation:

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

Given that NaH2PO4 is a weak acid, we can assume it is the acid (HA), and Na2HPO4 is the conjugate base (A-).

The pKa of NaH2PO4 can be calculated using the equation:

pKa = -log(Ka)

Where Ka is the acid dissociation constant. The Ka value for NaH2PO4 is 7.5 x 10^-8.

So, the pKa is:

pKa = -log(7.5 x 10^-8) = 7.12

Now, we can rearrange the Henderson-Hasselbalch equation to solve for [A-]/[HA]:

[A-]/[HA] = 10^(pH - pKa)

[A-]/[HA] = 10^(7.50 - 7.12) = 2.57

This means that the ratio of the concentration of the conjugate base to the concentration of the acid in the buffer should be 2.57.

Since we have 200 mL of the acid (HA) at a concentration of 0.10 M, we can calculate the concentration of the conjugate base (A-) as:

[A-] = [HA] * 2.57 = 0.10 M * 2.57 = 0.257 M

To calculate the amount of NaOH needed to make the solution, we can use the equation:

moles = concentration * volume

The volume of NaOH needed can be calculated as:

volume = moles / concentration

The concentration of NaOH is 1.0 M, and we need enough to react with the 200 mL of the conjugate base (A-) at a concentration of 0.257 M.

moles of NaOH = concentration * volume = 0.257 M * 0.200 L = 0.0514 moles

volume of NaOH = moles / concentration = 0.0514 moles / 1.0 M = 0.0514 L

Finally, we need to convert the volume from liters to milliliters:

volume of NaOH = 0.0514 L * 1000 mL/L = 51.4 mL

Therefore, we need to add 51.4 mL of 1.0 M NaOH to 200 mL of 0.10 M NaH2PO4 to make a buffer solution with a pH of 7.50.

To solve this question, we need to understand the principles of buffer solutions and how they work.

A buffer solution is a solution that can resist changes in pH when small amounts of acid or base are added to it. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, we have a weak acid, NaH2PO4, and its conjugate base, Na2HPO4.

The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffer solution:

pH = pKa + log ([Salt]/[Acid])

Where pKa is the negative logarithm (base 10) of the acid dissociation constant, [Salt] is the concentration of the conjugate base, and [Acid] is the concentration of the weak acid.

Given that we want to make a buffer solution with a pH of 7.50, we can rearrange the Henderson-Hasselbalch equation:

7.50 = pKa + log ([Salt]/[Acid])

Since we have the concentrations of the weak acid and the conjugate base, we can calculate the ratio [Salt]/[Acid].

Now, let's find the pKa value of the weak acid, NaH2PO4. The pKa value can be obtained from a reference source or calculated using the acid dissociation constant (Ka) equation:

ka = [Salt][H+]/[Acid]

We can assume that the dissociation of NaH2PO4 into Na+ and H2PO4- is complete, so the concentration of H2PO4- is equal to the initial concentration of NaH2PO4.

Using the given concentrations, we get:

0.10 M = [Salt][H+]/[Acid]

0.10 M/[H2PO4-] = [Salt]

Now let's substitute these values into the Henderson-Hasselbalch equation:

7.50 = pKa + log (0.10 M/[H2PO4-])

We can rearrange the equation to solve for pKa:

pKa = 7.50 - log (0.10 M/[H2PO4-])

Once we know the pKa value, we can use it to calculate the ratio [Salt]/[Acid]. In this case, the acid is NaH2PO4 and the salt is Na2HPO4.

Now let's calculate the ratio:

[Salt]/[Acid] = 10^(pH - pKa)

Substituting the given values:

[Salt]/[Acid] = 10^(7.50 - pKa)

Now we have the ratio [Salt]/[Acid].

To calculate the number of milliliters of 1.0 M NaOH needed, we need to consider that NaOH is a strong base that will react with the weak acid, NaH2PO4, to form water and the corresponding salt.

The reaction equation is:

NaH2PO4 + NaOH -> H2O + Na2HPO4

We can use the balanced stoichiometry of the reaction to calculate the moles of NaOH needed to react with the moles of NaH2PO4.

First, calculate the moles of NaH2PO4 using the given concentration (0.10 M) and volume (200 mL):

moles of NaH2PO4 = (0.10 M) x (0.200 L)

Next, calculate the moles of NaOH needed. Since the reaction has a 1:1 stoichiometric ratio between NaH2PO4 and NaOH, the moles of NaOH needed will be the same as the moles of NaH2PO4:

moles of NaOH = moles of NaH2PO4

Finally, calculate the volume of 1.0 M NaOH needed using the moles of NaOH and its concentration (1.0 M):

volume of NaOH = (moles of NaOH) / (1.0 M)

Now you can perform the calculations to obtain the final answer.