A buffer solution is prepared by mixing 1 mole of HA and 1 mole of NaA into 1 L distilled water. Calculate the change in pH when 2t mL of 0.20 M NaOH is added into 500mL of the buffer?

Answer 1

Here's what I got.

SIDE NOTE: I'll use an example of a sample of sodium hydroxide solution since you typed incorrectly the amount of strong base added to the buffer.

So, you're dealing with a buffer solution that contains a weak acid, #"HA"#, and its conjugate base, #"A"^(-)#, delivered to the solution by its salt, #"NaA"#.

As you may know, the Henderson-Hasselbalch equation can be used to determine the pH of a buffer solution that contains a weak acid and its conjugate base.

#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))" "#

This is what you have

#color(blue)(pK_a = - log(K_a))" "#, where
#K_a# - the acid dissociation constant for the weak acid
Since the problem doesn't provide you with a value for #K_a#, you will have to find a way to express the change in pH, #Delta_(pH)#, without using the #pK_a# of the acid.
So, you prepare your buffer solution by mixing #1# mole of weak acid and #1# mole of conjugate base.
SIDE NOTE The salt of the conjugate base dissociates in a #1:1# mole ratio with the conjugate base, so any number of moles of salt you have will be equal to the number of moles of conjugate base produced in solution.

In the stock solution, the two chemical species' molarities will be

#color(blue)(c = n/V)#
#["HA"] = ["A"^(-)] = "1 mole"/"1 L" = "1 M"#
Now, you're going to take a #"500-mL"# sample of this stock solution, This sample will contain
#c = n/V implies n = c * V#
#n_(HA) = n_(A^(-)) = "1 M" * 500 * 10^(-3)"L" = "0.50 moles"#

between a conjugate base and a weak acid.

Before adding the strong base, the pH of the buffer will be equal to the acid's #pK_a#, since you're dealing with equal concentrations of weak acid and conjugate base
#"pH"_1 = pK_a + overbrace(log( (1 color(red)(cancel(color(black)("M"))))/(1color(red)(cancel(color(black)("M"))))))^(color(purple)(=0))#
Let's assume that you're adding #"25 mL"# of sodium hydroxide solution. The hydroxide anions will react in a #1:1# mole ratio with the weak acid (neutralization reaction) and produce conjugate base
#"HA"_text((aq]) + "OH"_text((aq])^(-) -> "A"_text((Aq])^(-) + "H"_2"O"_text((l])#
The number of moles of hydroxide anions added to the #"500-mL"# buffer will be
#n_(OH^(-)) = "0.20 M" * 25 * 10^(-3)"L" = "0.0050 moles OH"^(-)#

Following the neutralization reaction, you will be left with

#n_(OH^(-)) = "0 moles" -># completely consumed
#n_(HA) = "0.50 moles" - "0.0050 moles" = "0.495 moles"#
#n_(A^(-)) = "0.50 moles" + "0.0050 moles" = "0.505 moles"#

At this point, the buffer's total volume will be

#V_"total" = "500 mL" + "25 mL" = "525 mL"#

The conjugate base and weak acid will now have different concentrations that will be

#["HA"] = "0.495 moles"/(525 * 10^(-3)"L") = "0.9429 M"#
#["A"^(-)] = "0.505 moles"/(525 * 10^(-3)"L") = "0.9619 M"#

At this point, the solution's pH will be

#"pH"_2 = pK_a + log( (0.9619 color(red)(cancel(color(black)("M"))))/(0.9429color(red)(cancel(color(black)("M")))))#
#"pH"_2 = pK_a + 0.00866#

As a result, the pH change will be

#Delta_(pH) = "pH"_2 - "pH"_1#
#Delta_(pH) = color(red)(cancel(color(black)(pK_a))) + 0.00866 - color(red)(cancel(color(black)(pK_a)))#
#Delta_(pH) = color(green)(0.00866)#

Repeat this process using the actual amount of sodium hydroxide that was provided to you in the problem.

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Answer 2

To solve this problem, you can use the Henderson-Hasselbalch equation:

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

First, calculate the initial concentrations of HA and A- in the buffer solution: [HA] = 1 M [A-] = 1 M

Next, calculate the initial pH using the equation: pH = pKa + log([A-]/[HA])

Given that HA is a weak acid, you need to find the pKa from a table or calculate it if it's provided. Let's assume the pKa value for HA is 4.

pH = 4 + log(1/1) = 4

After adding NaOH, you need to calculate the new concentrations of A- and HA. Since NaOH is a strong base, it completely dissociates to form Na+ and OH- ions.

Calculate the moles of NaOH added: moles NaOH = (0.20 mol/L) * (2 mL / 1000 mL) = 0.0004 moles

The buffer solution has a total volume of 500 mL, so it contains 0.5 moles of HA and 0.5 moles of A-.

After adding NaOH, the concentration of A- will increase by 0.0004 moles, and the concentration of HA will decrease by the same amount.

Calculate the new concentrations: [HA] = (0.5 moles - 0.0004 moles) / 0.5 L [A-] = (0.5 moles + 0.0004 moles) / 0.5 L

Now, plug these values into the Henderson-Hasselbalch equation to find the new pH:

pH = 4 + log([A-]/[HA])

Calculate [A-]/[HA]: [A-]/[HA] = (0.5 + 0.0004) / (0.5 - 0.0004)

Calculate pH using the equation: pH = 4 + log([A-]/[HA])

Finally, subtract the initial pH from the new pH to find the change in pH.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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