Chemistry, 28.07.2019 07:00, elysabrina6697

Calculate the ph after 0.020 mol hcl is added to 1.00 l of each of the four solutions below.

Answers: 1

Show answers

Answers

answered: Jacques1b

The pH of the solutions after adding 0.02mol of HCl will be 1.7, 1.69, 1.7, and 6.4 respectively.

pH is used to measure the alkalinity and acidity of the solutions. pH 7 is neutral.

For solution A,

$\bold {Kb =\dfrac {[OH^-][HONH3+]}{[NH2OH]} }$

put the value and solve it for [OH]

$\bold {[OH^-] = 6.6x10^-^5 M }$

$\bold{ pOH = -log[OH]}\\\\\bold{ pOH = -log(6.6x10^-^5)}\\\\\bold {pOH = 4.18}}\\$

Since, pH = 14 - pOH

pH = 9.82

Since, $\bold{[H^+] = 0.02}$

$\bold {pH = -log (0.02)}\\\bold {pH = 1.7}$

For Solution B

$\bold {Ka [HONH_3Cl ]= 9.09x10^-7 }\\\\\bold {Ka = \dfrac { [H+][HONH2]} { [HONH3+]}}$

Since, $\bold { [H^+] = 6x10^-^4 }$

So, pH = $\bold {pH = -log (6x10^-4) = 3.22}$

The new pH will be

$\bold {pH = -log (6x10^-4+0.02)}\\\bold {pH = 1.69}$

$\bold{[H^+] = 0.02}\\\bold {pH = -log (0.02)}\\\bold {pH = 1.7}$

(D) Since, $\bold {HONH_2}$ and $\bold {HONH_3Cl}$ are at same concentration.

So, $\bold {pH = pKa}$

$\bold {Pka = -log Ka}\\\\\bold {pKa = -log 9.09x10^-7}\\\\\bold{ pKa = 6.04}\\\\\bold{ pH = 6.04}$

Therefore, The pH of the solutions after adding 0.02mol of HCl will be 1.7, 1.69, 1.7, and 6.4 respectively.

To know more about pH

link

answered: meiyrarodriguez

A) according to this reaction:
by using ICE table:
NH2OH(aq) + H2O(l) → HONH3+(aq) + OH-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X

when Kb = [OH-][HONH3+]/[NH2OH]
when we have Kb = 1.1x10^-8 so,
by substitution:
1.1x10^-8 = X^2/(0.4-X) by solving this equation for X

∴X = 6.6x10^-5 M
∴[OH] = 6.6x10^-5 M
when POH = - ㏒[OH]
∴POH = -㏒(6.6x10^-5)= 4.18
∴PH = 14 - POH = 14 - 4.18
= 9.82
when PH = -㏒[H+]
∴[H+] = 10^9.82 = 1.5x10^-10 M+0.02molHcl
= 0.02
∴ the new value of PH = -㏒(0.02)
∴PH = 1.7

B) according to this reaction:
by using ICE table:
HONH3+(aq) → H+(aq) + HONH2(aq)
intial 0.4 0 0
change -X +X +X
Equ (0.4-X) X X

when Ka HONH3Cl = 9.09x10^-7
and Ka = [H+][HONH2] / [HONH3+]

So by substitution and we can assume [HONH3+] = 0.4 as the value of Ka is so small so,
9.09x10^-7 = X^2 / 0.4 by solving for X
∴ X = 6 x 10 ^-4
∴[H+] = 6x10^-4
PH = -㏒[H+]
= -㏒ (6x10^-4) = 3.22
when [H+] = 6x10^-4 + 0.02 m HCl
∴new value of PH = -㏒(6x10^-4+0.02)
= 1.69
C) when we have pure H2O and PH of water = 7
So we can get [H+] when PH = -㏒[H+]
∴[H+] = 10^-7 + 0.02MHCl
= 0.02
∴new value of PH = -㏒0.02
PH = 1.7
d) when HONH2 & HONH3Cl have the same concentration and Hcl added to them so we can assume that PH=Pka
and when we have Ka for HONH3Cl = 9.09x10^-7
So we can get the Pka:

Pka = -㏒Ka
= -㏒9.09x10^-7
= 6.04
∴PH = 6.04
and because of the concentration of the buffer components, HONH2 & HONH3Cl have 0.4 M and the adding of HCl = 0.02 M So PH will remain very near to 6

answered: Guest

What do you mean by that? like the contents? the thc level?

answered: Guest

Less stable hope this

Other questions on the subject: Chemistry

Chemistry, 22.06.2019 01:20, sarinaneedshelp01

Match the acid base pairs by arranging the acid name with the conjugate base formula. hydrogen carbonate hydrogen phosphate carbonic acid read water sulfuric acid phosphoric acid a. co32- b. hso4- c. hco3- d. po43- e. h2po4- f. oh-

Answers: 1

continue

Chemistry, 22.06.2019 03:50, Pizzapegasus1

Express the following number in scientific notation. 0.026890 =

Answers: 1

continue

Chemistry, 22.06.2019 07:30, sjsmith21

What is i fracture in the crust called when land move up, down or sideways

Answers: 2

continue

Chemistry, 22.06.2019 15:30, lovebaeforlife351

Each of the following reactions is allowed to come to equilibrium and then the volume is changed as indicated. predict the effect (shift right, shift left, or no effect) of the indicated volume change. drag the appropriate items to their respective bins. co(g) + h2o(g) < => co2(g) + h2(g) (volume is decreased) pcl3(g) + cl2(g) < => pcl5(g) (volume is increased) caco3(s)< => cao(s) + co2(g) (volume is increased)

Answers: 1

continue

Do you know the correct answer?

Calculate the ph after 0.020 mol hcl is added to 1.00 l of each of the four solutions below....