A 25.0 mL sample of a 0.2100 M solution of aqueous trimethylamine is titrated with a 0.2625 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25°C.

Volume required for neutralization V will be:

V * 0.2125 M HCl = 25 mL * 0.17 M  

V = 20 ml

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Explanation:

 When 10 mL is added we can apply Henderson equation to get the result, so:

The pH will be of basic buffer

pOH = pKb + log(salt/base)

or pOH = 4.19 + log (0.2125*10 / 25*0.17 - 10*0.2125 )

pOH = 4.19 and pH = 14 - 4.19 = 9.81 

Second part:

When 20 ml is added, there is only salt formed

The pH will be salt of strong acid and weak base

So pH = 7 - 0.5 pKb - 0.5 log C

where C is the concentration of the salt formed so:

pH = 7 - (0.5*4.19) - (0.5 log (25*0.17) / (25+20))

  = 5.42

Third part:

When 30 ml of the acid has been added,

The pH will be of the remaining strong acid

pH = - log (0.2125*10 / 25 + 30 ) 

  = 1.326 

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answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e

1 a

2 d

3 a

4 c

explanation: