Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a strong acid; ka for hso4− = 1. 3 × 10−2. ) [hso4−] = m [so42−] = m [h ] = m

Sagot :

tushargupta0691

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid.  HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R   [tex]HSO_4^-[/tex]    ⇄  [tex]H^+ + SO_4^2^-[/tex]

I    [tex]0.14[/tex]

C   [tex]- x[/tex]               [tex]+x[/tex]       [tex]+x[/tex]

E   [tex]0.14-x[/tex]        [tex]x[/tex]         [tex]x[/tex]

[tex]K_a = 1.3[/tex] × [tex]10^-^2[/tex] for [tex]HSO^-_4[/tex] . As a result,

[tex]\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a[/tex]

[tex]K_a[/tex] is large. It is no longer valid to approximate that [tex][HSO^-_4][/tex] at equilibrium is the same as its initial value.

[tex]\frac{x^2}{0.14-y} = 1.3 * 10^-^2[/tex]

[tex]x^2+1.3*10^-^2x - 0.14[/tex] × [tex]1.3[/tex] × [tex]10^-^2= 0[/tex]

Solving the quadratic equation for [tex]x , x \geq 0[/tex] since [tex]x[/tex] represents a concentration;

                             [tex]x=0.0366538[/tex]

Then, round the results to 2 significant figure;

  • [tex][SO_4^2^-] = x = 0.037 mol. L ^-^1[/tex]
  • [tex][H^+] = x = 0.037 mol. L ^-^1[/tex]
  • [tex][HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1[/tex]

Learn more about concentration here:

https://brainly.com/question/14469428

#SPJ4

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.