Chemistry, 24.07.2019 17:00 keilahkimbrough8

A25.0 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.

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Answer from: joshlynn52

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 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.

Answer from: Quest

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

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A25.0 ml sample of a 0.1700 m solution of aqueous trimethylamine is titrated with a 0.2125 m solutio...

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