Anwer the following questions: -How many grams of ammonium chloride and how many

Question

Anwer the following questions:

-How many grams of ammonium chloride and how many milliliters of 3.0 M NaOH should be added to 200 mL of water and diluted to 500 mL to prepare a buffer solution of pH 9.50 with a salt concentration of 0.10 M? R = 7.49 g of NH 4 Cl, and 30 mL of NaOH.

-The pH of the blood is 7.40. What is the ratio of [HPO42 -] / [H2PO4-] in the blood at 25 ° C? R = 1.9 / 1

- What is the pH of a solution that is 0.20 M in phthalic acid (H2P) and 0.10 M in phthalate acid potassium (KHP)? R = 2.62

Explanation / Answer

1) pH=pka +log [base]/[acid] (Henderson-hasselbach)

pka (NH4+)=9.3

NH4+ +OH- --->NH3+H3O+

9.50=9.25+log [NH3]/[NH4+]

9.50-9.25=log [NH3]/[NH4+]

0.25=log [NH3]/[NH4+]

[NH3]/[NH4+]=10^0.25=1.778

So [NH4+]=0.1M

[NH3]=1.778*[NH4+]=1.778*0.1M=0.1778M

In 500 ml=0.5L of the buffer,mol of NH3=0.1778mol/L*0.5L=0.0889mol

This means 0.0889mol of NaOH needs to be added to neutralize NH4+ to NH3

So volume of NaOH added=0.0889mol/3.0mol/L=0.0296L=30ml(approx)

And mol of NH4Cl added initially=mol of NH3 +mol of NH4+ in the buffer=0.0889mol+0.1mol/L*0.5L=0.0889+0.05=0.1389mol

So mass of NH4Cl added initially=0.1389mol*53.481g/mol=7.429g

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