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To calculate the amount of HEPES-acid and 1M NaOH needed to make 500mL of a 0.3M HEPES buffer at pH 7.35, we can use the Henderson-Hasselbalch equation. The pKa of HEPES is 7.55. First, we need to determine the ratio of HEPES-acid to HEPES-base required to achieve a pH of 7.35. This can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) 7.35 = 7.55 + log([A-]/[HA]) -0.2 = log([A-]/[HA]) 10^(-0.2) = [A-]/[HA] [A-]/[HA] = 0.63 This means that for every 0.63 moles of HEPES-base, we need 1 mole of HEPES-acid. Since we want a 0.3M HEPES buffer in 500mL, we need 0.3 moles of HEPES in total. Let x be the amount of HEPES-base in moles. Then, the amount of HEPES-acid needed is 0.63x. x + 0.63x = 0.3 1.63x = 0.3 x = 0.1847 moles Therefore, we need 0.1847 moles of HEPES-base and 0.1162 moles of HEPES-acid to make 500mL of a 0.3M HEPES buffer at pH 7.35.

          To calculate the amount of HEPES-acid and 1M NaOH needed to make 500mL of a 0.3M HEPES buffer at pH 7.35, we can use the Henderson-Hasselbalch equation. The pKa of HEPES is 7.55.

First, we need to determine the ratio of HEPES-acid to HEPES-base required to achieve a pH of 7.35. This can be calculated using the Henderson-Hasselbalch equation:

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

7.35 = 7.55 + log([A-]/[HA])

-0.2 = log([A-]/[HA])

10^(-0.2) = [A-]/[HA]

[A-]/[HA] = 0.63

This means that for every 0.63 moles of HEPES-base, we need 1 mole of HEPES-acid.

Since we want a 0.3M HEPES buffer in 500mL, we need 0.3 moles of HEPES in total.

Let x be the amount of HEPES-base in moles. Then, the amount of HEPES-acid needed is 0.63x.

x + 0.63x = 0.3

1.63x = 0.3

x = 0.1847 moles

Therefore, we need 0.1847 moles of HEPES-base and 0.1162 moles of HEPES-acid to make 500mL of a 0.3M HEPES buffer at pH 7.35.

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Added by Jay C.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Dominique Jan Tan

08/08/2023

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Given that x represents the amount of HEPES-base in moles, we have: x + 0.63x = 0.3 1.63x = 0.3 x = 0.3 / 1.63 x = 0.1847 moles Show more…

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To calculate the amount of HEPES-acid and 1M NaOH needed to make 500mL of a 0.3M HEPES buffer at pH 7.35, we can use the Henderson-Hasselbalch equation. The pKa of HEPES is 7.55. First, we need to determine the ratio of HEPES-acid to HEPES-base required to achieve a pH of 7.35. This can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) 7.35 = 7.55 + log([A-]/[HA]) -0.2 = log([A-]/[HA]) 10^(-0.2) = [A-]/[HA] [A-]/[HA] = 0.63 This means that for every 0.63 moles of HEPES-base, we need 1 mole of HEPES-acid. Since we want a 0.3M HEPES buffer in 500mL, we need 0.3 moles of HEPES in total. Let x be the amount of HEPES-base in moles. Then, the amount of HEPES-acid needed is 0.63x. x + 0.63x = 0.3 1.63x = 0.3 x = 0.1847 moles Therefore, we need 0.1847 moles of HEPES-base and 0.1162 moles of HEPES-acid to make 500mL of a 0.3M HEPES buffer at pH 7.35.

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Transcript

00:01 A buffer is a solution that is formed from the combination of a weak acid or a weak base together with its salt.

00:12 For the combination of a weak acid and its salt, we will use the henderson -hasselbalch equation in order to solve for the ph in this problem.

00:24 We write it as ph equals pka plus the log of the moles of the salt which is sodium acetate that's nac2h2o2 divided by the moles of the acid which is acetic acid that is hc2h3o2.

00:44 So note that a solution that is a buffered solution contains both acid and base components.

00:53 Let us first compute for the moles of the acid and the salt using the idea that we have moles equals molarity times the volume.

01:06 If we wish to find the minimums instead, we will multiply the concentration by the ml.

01:15 So once again, molarity times ml.

01:19 So the molarity of sodium acetate, we have sodium acetate is 0 .0265.

01:26 Volume of the solution is 480.

01:29 So with this, we get 12 .72.

01:35 For the acid itself, minimums, the concentration of the acetic acid is 0 .0185.

01:48 The volume is 480.

01:51 And this gives us 8 .83.

01:58 Note that if base is added, naoh is added.

02:12 When base is added, the amount of the acid will react, the acid will react with the base so that the amount of the acetic acid will decrease by the amount of the base added.

02:27 So let's say first that the minimals of naoh added is equal to its molarity, which is 5 .80 times the volume x which we will be looking for in this problem.

02:41 So once again, the minimals of the acid will decrease by the amount of the base we added.

02:48 So the initial moles of 8 .88 minus 5 .80x while the moles of the salt will increase by the minimals of the base added.

03:01 We have 12 .72 plus 5 .80 times x...

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