Question

A 25.0 ml sample of a 0.1700 m solution of aqueous trimethylamine is titrated with a 0.2125 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.

Answer 1

Volume required for neutralization V will be:
V * 0.2125 M HCl = 25 mL * 0.17 M 
V = 20 ml 
First part:
 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 

Answer 2

Answer:

For 10.0 mL, pH=9.81

For 20.0 mL, pH=5.42

For 30.0 mL, pH=1.41

Explanation:

Hello,

At first, the undergoing chemical reaction is:

$HCl+(CH_3)_3N-->(CH_3)_3NH^++Cl^-$

Nevertheless, once the acid added but before the equivalent point, the trimethylamine reacts with the dissociated hydrogen ions provided by the hydrochloric acid as shown below:

$(CH_3)_3N+H^+(CH_3)_3NH^+$

A) 10 mL of HCl

In this manner, the fist addition of 10 mL HCl leads to the following pH calculation:

Before the reaction:

$n_{(CH_3)_3N}=0.0250L*0.1700mol/L=0.00425mol\\n_{H^+}=0.01L*0.2125mol/L=0.002125mol$

After the reaction:

$M_{(CH_3)_3N}=\frac{0.00425mol-0.002125mol}{0.025L+0.01L} =0.061M\\M_{H^+}=\frac{0.002125mol-0.002125mol}{0.025L+0.01L} =0M\\M_{(CH_3)_3NH^+}=\frac{0.002125mol}{0.025L+0.01L} =0.061M$

Now, the conversely chemical reaction ($(CH_3)_3NH^+(CH_3)_3N+H^+$) defines the pH via the equilibrium law of mass action calculation:

$Ka=\frac{[H^+]_{eq}[(CH_3)_3N]_{eq}}{[(CH_3)_3NH^+]_{eq}}$

Ka is computed via:

$Ka=\frac{Kw}{Kb}=\frac{1x10^{-14}}{10^{-4.19}}=1.54x10^{-10}$

Now, defining $x$ as the concentration change during the chemical reaction, its value will account for the concentration of hydrogen which subsequently allows us to compute the pH considering that the initial concentrations match with the concentrations after the addition of the 10mL of HCl, thus:

$1.55x10^{-10}=\frac{x*(0.061M+x)}{0.061M-x} \\solving\ for\ x=1.55x10^{-10}M\\pH=-log(1.55x10^{-10})=9.81$

B) Now, after 20 mL have been added we obtain:

Before the reaction:

$n_{(CH_3)_3N}=0.0250L*0.1700mol/L=0.00425mol\\n_{H^+}=0.02L*0.2125mol/L=0.00425mol$

After the reaction:

$M_{(CH_3)_3N}=\frac{0.00425mol-0.00425mol}{0.025L+0.02L} =0M\\M_{H^+}=\frac{0.00425mol-0.00425mol}{0.025L+0.02L} =0M\\M_{(CH_3)_3NH^+}=\frac{0.00425mol}{0.025L+0.02L} =0.0944M$

By applying the same previous procedure, the pH is then:

$1.55x10^{-10}=\frac{x*x}{0.094M-x} \\solving\ for\ x=3.82x10^{-6}M\\pH=-log(3.82x10^{-6})=5.42$

(here is the equivalence point).

C) Finally, after 30.0mL have added:

Before the reaction:

$n_{(CH_3)_3N}=0.0250L*0.1700mol/L=0.00425mol\\n_{H^+}=0.03L*0.2125mol/L=0.006375mol$

After the reaction:

$M_{(CH_3)_3N}=\frac{0.00425mol-0.00425mol}{0.025L+0.03L} =0M\\M_{H^+}=\frac{0.006375mol-0.00425mol}{0.025L+0.03L} =0.0386MIn this case, the concentration of hydrogen in excess, 0.0386M is just enough to compute the pH since all of the base is consumed, thus:[tex]pH=-log(0.0386)=1.41$

Best regards.