given the information that the pH of the acetic acid/acetate buffer= 5.1, how wo

Question

given the information that the pH of the acetic acid/acetate buffer= 5.1, how would you calculate the amount of salts you need in order to prepare the buffer? (question 1)

community.acphs.edu ] Blackboard Learn https://community.acphs.edu/bbcswebdav/pid-506608-dt-content-rid-3602383_2/courses/2016SPCHE1210BB/buff... buffer. Part 2: Prepare a Phosphate Buffer 1. Calculate the amount of each salt you will need to make up your assigned buffer. You will be using two of the salts of phosphoric acid (a polyprotic acid): anhydrous KH2PO4 and K2HPO4 3H20. The H2PO4 ion is the weak acid (K 6.2 x 10-8) and the HPO42 ion is the conjugate base. Make sure that you are going to weigh out reasonable amounts of each salt (about 0.5-3.0 g or so). HINT: Use the Henderson-Hasselbalch equation and solve for the ratio of the base/acid in terms of Ka and [H']. You know the Ka and the pH, so you can find the [H']. Pick a molarity for one of the components (either [base] or [acid]) and calculate the other. Once you have the concentrations of the two salts (that is molarity) you can find the masses of the salts needed to prepare your buffer. If you get masses that are too big, divide each mass by the same number; the ratio stays the same. If you get masses that are too small, multiply each mass by the same number. Prepare the phosphate buffer in the volumetric flask provided. HINT: Put the solids into the flask and dissolve them with some deionized water. Dilute to the mark with more DI water. 2.

Explanation / Answer

For buffer,

pH = pKa + log(base/acid)

let the total concentration of buffer be 1.0 M and total volume be 1 L

Given pH = 5.1

pKa = 4.74

we get,

5.1 = 4.74 + log([CH3COO-]/[CH3COOH])

[CH3COO-] = 2.29[CH3COOH]

we assumed,

[CH3COOH] + [CH3COO-] = 1 M

[CH3COOH] = 1 - [CH3COO-]

Feed into above equation,

[CH3COO-] = 2.29(1 - [CH3COO-]

[CH3COO-] = 2.29 - 2.29[CH3COO-]

So,

concentration of [CH3COO-] required = 0.70 M

mass of CH3COONa required = 0.70 M x 1 L x 82.03 g/mol = 57.421 g

concentration of [CH3COOH] required = 0.30 M

mass of CH3COOH required = 0.3 M x 1 L x 60.05 g/mol = 18.015 g

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