All categories

3 years ago

What is the pH of a solution with a concentration of 4.2 x 10−5 M H3O+? (4 points)

Chemistry

1 answer:

zheka24 [161]3 years ago

4 0

Answer:

1) b. 4.38

2) d. 10.7

Explanation:

1) The problem tells us that [H₃O⁺] = 4.2x10⁻⁵ M, and keeping in mind that [H₃O⁺]= [H⁺], we can calculate the pH of the solution:

pH = -log[H⁺] = 4.38

2) First we calculate the pOH of the solution:

pOH = -log[OH⁻] = -log(4.6x10⁻⁴) = 3.33

Then we calculate the pH using the following formula:

pH = 14 - pOH = 14 - 3.33

pH = 10.7

You might be interested in

For the following reaction at 600. K, the equilibrium constant, Kp, is 11.5. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g)

Lubov Fominskaja [6]

Answer:

a) pPCl5 = 0.856 atm

b)pPCl5 = 0.0557 atm

pCl2 = pCl3 = 0.800 atm

c) Ptotal = 1.66 atm

d) 93.5

Explanation:

Step 1: Data given

Temperature = 600 K

Kp = 11.5

Mass of PCl5 = 2.010 grams

Volume of the bulb = 555 mL = 0.555 L

The bulb is heated to 600 K

Step 2: The balanced equation

PCl5(g) ⇄ PCl3(g) + Cl2(g)

Step 3:

a) pv = nrt

⇒with p = the pressure = TO BE DETERMINED

⇒with V = the volume = 0.555 L

⇒ with n =the number of moles PCl5 = 2.010 grams / 208.24 g/mol = 0.00965 moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 600K

p = (0.00965 *0.08206*600)/0.555

pPCl5 = 0.856 atm

The initial pressures

pPCl5 = 0.856 atm

pCl2 = pCl3 = 0 atm

For 1 mol PCl5 we'll have PCl3 and 1 mol Cl2

The pressure at the equilibrium

pPCl5 = (0.856 -x) atm

pCl2 = pCl3 = x atm

Kp = pPCl3 * pCl2/pPCl5

11.5 = x*x / (0.856 - x)

11.5 = x²/(0.856- x)

x = 0.8003

pPCl5 = (0.856 -x) atm = 0.0557 atm

pCl2 = pCl3 = x atm = 0.800 atm

c) Since x = 0.8003 and PCl3 and PCl2 are x

Ptotal = 0.8003 + 0.8003 +0.0557 = 1.66 atm

The degree of dissociation = (x / initial pressure PCl5)

(0.8003/0.856) * 100 = 93.5

7 0

4 years ago

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

melisa1 [442]

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

4 0

3 years ago

What country was the element hydrogen discovered in

alexdok [17]

I believe it was discovered in france

5 0

3 years ago

How many unpaired electrons are there in helium?

Crazy boy [7]

There are 2. Helium is composed of two electrons bound by the electromagnetic force to a nucleus containing two protons along with either one or two neutrons, depending on the isotope, held together by the strong force.

7 0

4 years ago

Read 2 more answers

Identify whether each compound is a straight chain, branched chain, or cycloalkane,

Delvig [45]

Answer:

First Part: branched chain, cycloalkane, straight chain

Second Part: straight chain, branched chain, cycloalkane.

Explanation:

Edg 2021

7 0

3 years ago