A common practice in early stages of research is to simulate the conditions of b

Question

A common practice in early stages of research is to simulate the conditions of blood using a simple buffered system. You are tasked with the responsibility to prepare 250 mL of buffer with a total concentration on 125 millimolar(mM) at a pH of 6.85 as a matrix representing blood suffering from a slight case of acidosis.

(a) Of the three options given below, select the best acid as a starting point for the preparation of this buffer in one sentence explianing your reasoning

(b) Calculate the mass (in milligrams) of the orginal solid you selected in part (a) and the volume of a solution of 6.0 M NaOH (in mL) that would be requiored to prepare the desired buffer?

(c) give one specirfic potential source of error that could have occured during the preparation of the buffer and explian in detail how you would expect it to affect either the final pH or overall concentration of the buffer?

Explanation / Answer

(a) Of the three options given below, select the best acid as a starting point for the preparation of this buffer in one sentence explaining your reasoning

a(HOCH2)3CH3Cl 8.51 x 10-9CH3CO2H 1.74 x 10-5NaH2PO46.31 x 10-8

You should chooose and acid with the pKa in the range of the desired pH

a) pKa = - log Ka = - log 8.51 x 10-9=  8.07

b) pKa = - log Ka = - log 1.74 x 10-5=  4.76

c) pKa = - log Ka = - log 6.31 x 10-8=  7.20

Since the desired pH is 6.85 the closest pKa is for NaH2PO4so we will choose this

pH = pKa + log base/acid

6.85 = 7.2 log base/acid

log base/acid = - 0.35

base/acid = 10-0.35

base/acid = 0.447

We need 250 mL of 125 mM concentration so 0.25 L x 0.125 M = 0.03125 moles

NaH2PO4 MW = 120 g/mol

0.03125 moles will be 0.03125 x 120 = 3.75 g of 3750 mg

since base/acid = 0.447

base + acid = 0.03125

acid = 0.03125 -base

base/0.03125 -base = 0.447

base = 0.01397 - 0.447base

1.447base = 0.01397

base = 0.01397/1.447 = 0.00965 moles

since we have 6 M NaOH

6 M x V = 0.00965

V = 0.00965/6 = 0.0016 = 1.6 mL 6 M NaOH

So you will need 1.6 mL 6 M NaOH and 3750 mg NaH2PO4 in 250 mL to prepare the required buffer.

3) Since the amount of NaOH to be added is small and the concentration is high there can be small error in adding this volume of the base, if you add more the pH will be higher and if you add slightly less the pH will be lower than the calculated value but sice it is a buffer the pH difference will be negligible.

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