Chemistry, 10.03.2020 09:06 amh926292

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the amino acid alanine are 2.344 ( p K a1 ) and 9.868 ( p K a2 ) , which correspond to the carboxylic acid and amino groups, respectively. a) how do you calculate the PH of the first equivalent and b) the second equivalent? please help

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Answer from: rchapman414

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

H_2NC_2H_5CO^-_2 H^+ ------> H_3}^+NC_2H_5CO^-_2

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

[H_3}^+NC_2H_5CO^-_2] = $\frac{initial moles of alaninate}{total volume}$

[H_3}^+NC_2H_5CO^-_2] = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

[H_3}^+NC_2H_5CO^-_2] = $\frac{8}{100mL}$

[H_3}^+NC_2H_5CO^-_2] = 0.08 M

Alanine serves as an intermediary form, however the concentration of H^+ and the pH can be determined as follows;

[H^+] = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

[H^+] = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

[H^+] = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

[H^+] = $7.63*10^{-7}M$

pH = - log [H^+]

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

H_2NC_2H_5CO^-_2 H^+ -----> H^+_3NC_2H_5CO^-_2

H^+_3NC_2H_5CO^-_2 H^+ -----> H^+_3NC_2H_5CO_2H

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

[H^+_3NC_2H_5CO_2H] = $\frac{initial moles of alaninate}{total volume}$

[H^+_3NC_2H_5CO_2H] = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

[H^+_3NC_2H_5CO_2H] = $\frac{8}{150}$

[H^+_3NC_2H_5CO_2H] = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

H^+_3NC_2H_5CO_2H ⇄ H^+_3NC_2H_5CO^-_2 H^+

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

0.00453(0.053-x) =x^2

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So [H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M

pH = -log[H^+]

pH = -log[0.0134]

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

Answer from: Quest

they are called fossil fuels

Answer from: Quest

The reaction 2kalsi3o8 + 2(h+ + hco3-) + h2o → al2si2o5(oh)4 + 2k+ + 2hco3- + 4sio2 shows that feldspar kalsi3o8 is chemically altered by hydrolysis to form chalky-looking clay mineral kaolinite al2si2o5(oh)4. the chemical breakdown of potassium feldspar due to exposure to acidic water results to the mineral binding with the water molecule to become kaolinite with silica and potassium and bicarbonate ions in solution.

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50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl...