100 mL of 0.5M NaCl in 25 mM sodium phosphate, pH 8.0. Provide calcualtions and explain...

100 mL of 0.5M NaCl in 25 mM sodium phosphate, pH 8.0.

Provide calcualtions and explain how the 0.5M NaCl in 25 mM sodium phosphate buffer, pH 8, will be prepared.

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Expert Solution

Using the Henderson equation, pH = pKa + log [Base]/[acid]

NaH2PO4 as acid (119.96 g/mol) and Na2HPO4 as base (141.96 g/mol) will be used to make the buffer. the pKa reported for this buffer is 7.21 so:

now, 8= 7.21 +log(B/A) solving for B/A relation:

B/A = 0.1 ---> B = 0.1A

25mM, 100 mL sodium phosphate bugffer = 0.1 L x 0.025 mol/L = 2.5x10^-3 mol

Buffer = B + A and we know that B = 0.1 A so:

0.025 = 0.1A + A solving for A:

1.1 A = 0.025

A = 0.0227 M * 0.1 L = 0.00227 moles of A

m of A = 0.00227 * 119.96 = 0.272 g

m of B = 0.0025 * 141.96 = 0.3549 g

mix these two quantities in 100 mL of water to make 25mM sodium phosphate buffer solution,

now, 100 mL of 0.5 M NaCl:

moles = 0.1 * 0.5 = 0,05 moles of NaCl

m of salt = 0,05 * 58.44 = 2.92 g,

mix this amont of NaCl to 100 mL of 25mM sodium phosphate buffer solution to obtain the final solution. with a pH of 8.

If you need more explanation, something to be fixed or still has doubt tell me in a comment and I'll gladly help you out.

Hope this helps

queen_honey_blossom answered 1 year ago

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