Can you please solve this question with all parts. Thank You. Step by step solut

Question

Can you please solve this question with all parts. Thank You. Step by step solution

A 50 mL of water requires 12 mL of 0.02-N H_2SO_4 to titrate it to the phenolphthalein end point and require 16mL of 0.02-N H_2SO_4 to titrate it further to the methyl-orange endpoint. What is the bicarbonate alkalinity in mg/L as CaCO_3? What is the carbonate alkalinity in mg Las CaCO_3? What is the total alkalinity in mg/L as CaCO_3? What is (are) the major alkalinity species and its (their) concentration(s) in meq/L? What is (are) the major acidity species and its (their) concentration(s) in meq/L?

Explanation / Answer

Here, P=12 and M=16

Condition : P > ½ M, Sample contains OH- and CO32- only.

1) So, the bicarbonate alkalinity is 0 mg/L, since the sample contains only OH- and CO32-.

2) For carbonate alkalinity, the formula is 2[M-P].

=2[16-12] = 8ml.

V1= Vol. of acid= 8ml

V2= Vol. of water sample= 50ml

N1= Normality of acid= 0.02 N

N2= Normality of water= ?

N2 = V1N1/V2 = (8*0.02)/50 = 0.0032 N

Amount of Carbonate alkalinity in terms of CaCO3= N2* Eq. weight of CaCO3 = 0.0032*50 gms = 0.0032*50*1000 mg/L = 160 mg/L

3) Total alkalinity= OH- and CO32- alkalinity

OH- alkalinity=

2P-M = 2*12-16 = 8 ml

For, OH- alkalinity also, N2 = 0.0032N

So, Amount of OH- alkalinity in terms of CaCO3= N2* Eq. weight of CaCO3 = 0.0032*50 gms = 0.0032*50*1000 mg/L = 160 mg/L

So, total Alkalinity= 160+160 mg/L= 320 mg/L

4) Major alkalinity species are OH- and CO32- . Their concentrations are 3.2 meq/L and 3.2 meq/L respectively.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---