Chemistry, 22.11.2019 06:31 myaaa13754

Find the concentration of h+ ions at a ph = 11 and
ph = 6. then divide the concentration of h+ ions at a
ph = 11 by the of h+ ions at a ph = 6. record your answer in table c.

what is the concentration of h+ ions at a ph = 11?

mol/l

what is the concentration of h+ ions at a ph = 6?

mol/l

how many fewer h+ ions are there in a solution at a
ph = 11 than in a solution at a ph = 6?

Answers: 2

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Answers

Answer from: izzybellee20004

Explanation:

the pH of the solution defined as negatuve logarithm of H^+ ion concentration.

$pH=-\log[H^+]$

1. Hydrogen ion concentration when pH of the solution is 11.

$11=-\log[H^+]$

$[H^+]=1\times 10^{-11} mol/L$ ..(1)

At pH = 11, the concentration of H^+ ions is $1\times 10^{-11} mol/L$ .

2. Hydrogen ion concentration when the pH of the solution is 6.

$6=-\log[H^+]'$

$[H^+]'=1\times 10^{-6} mol/L$ ..(2)

At pH = 6, the concentration of H^+ ions is $1\times 10^{-6} mol/L$ .

3. On dividing (1) by (2).

$\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} mol/L}{1\times 10^{-6} mol/L}=1\times 10^{-5}$

The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is $1\times 10^{-5}$ .

4. Difference between the H^+ ions at both pH:

$1\times 10^{-6} mol/L-1\times 10^{-11} mol/L=9.99\time 10^{-7} mol/L$

This means that Hydrogen ions in a solution at pH = 7 has $9.99\time 10^{-7} mol/L$ ions fewer than in a solution at a pH = 6

Answer from: etampus0220

for table C

Explanation:

What is the concentration of H+ ions at a pH = 11?

1. ⇒ 0.00000000001 mol/L

What is the concentration of H+ ions at a pH = 6?

2. ⇒ 0.000001 mol/L

How many fewer H+ ions are there in a solution at a

pH = 11 than in a solution at a pH = 6?

3. ⇒ 100,000

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.

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