Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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 :

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 It is no longer valid to approximate that [tex][HSO^-_4][/tex] at equilibrium is the same as its initial value.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 appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.