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Nataly_w [17]
3 years ago
14

2mL of a serum sample was added to 18mL of phosphate buffered saline (PBS) in Tube 1. 10mL of Tube 1 was added to 40mL of PBS in

Tube What is the dilution of serum in Tube 2?
Chemistry
1 answer:
Sladkaya [172]3 years ago
4 0

Answer:

Tube 2 has a total dilution of 1:50

Explanation:

We have a 2 ml serum sample added to a 18 mL phosphate buffered saline sample in tube 1. This means now in tube 1 there is 20 mL.

We have a 1:10 (= 2:20) dilution here.

10 ml of this 1:10 diluted tube 1 is taken and added to a 40 mL of PBS in tube 2.

Now we have 50 mL in tube 2.

This is a 10:50 (= 1:5) dilution.

The total dilution is 10x5 = 50

So the total ditultion has a rate 1:50

Tube 2 has a total dilution of 1:50

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For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
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Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

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