Chemistry, 08.06.2021 03:20 avisconti571

A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.950 M NaOH. The acid required 27.4 mL of base to reach the equivalence point. After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid?

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Answer from: Jenniferlam12

Solution :

The equation is :

$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$

The number of the moles of HA os 0.00285, and the volume is 25 mL.

15 mL of the 0.0950 M NaOH is added.

The total volume of a solution is V = 25 mL + 15 mL = 40 mL

The pH of the solution is 6.50

Calculating the K_a of HA

$HA(aq) \rightleftharpoons A^-(aq)+H^+$

$K_a=\frac{[A^-].[H^+]}{[HA]}$

Let s calculate the concentration of HA and NaOH

$[HA] = \frac{^nH_A}{V}$

$=\frac{0.00285 \ mol}{0.04 \ L}$

= 0.07125 M

$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$

$=\frac{0.001425 mol}{0.04L}$

= 0.0356 M

$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$

Initial conc. (M) 0.07125 M 0.0356 M 0 M

Change in conc. (M) -0.0356 M -0.0356 M + 0.0356 M

Equilibrium conc. (M) 0.03565 M 0 M 0.0356 M

Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M

0.0356 M of NaA dissociates completely into 0.0356 M Na^+ and 0.0356 M A^-

Now for [H^+]

$[H^+] = 10^{-pH}$

$=10^{-6.5}$

$=3.16 \times 10^{-7}$

Calculating the value of K_a ,

$K_a=\frac{[A^-].[H^+]}{[HA]}$

$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$

$=3.16\times 10^{-7}$

Therefore the the value of K_a for the unknown acid is $3.16\times 10^{-7}$ .

Answer from: Quest

n2(g) + 3h2-> 2nh3(g) this is the balanced equation

note the mole ratio between n2, h2 and nh3. it is 1 : 3 : 2 this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note: the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

Answer from: Quest

the food using pesticides in the future will harm you by giving you diseases like cancer or something since they have things that we should not eat in it like chemicals or pesticides. that is why organic food is more expensive because they take the time to make it right and healthy for you to eat.

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A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and ti...

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