When a deer eats too much in a given area, we call this ________

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 s

disa [49]

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 $HSO_4^-$ ⇄ $H^+ + SO_4^2^-$

I $0.14$

C $- x$ $+x$ $+x$

E $0.14-x$ $x$ $x$

$K_a = 1.3$ × $10^-^2$ for $HSO^-_4$ . As a result,

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

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 It is no longer valid to approximate that $[HSO^-_4]$ at equilibrium is the same as its initial value.

$\frac{x^2}{0.14-y} = 1.3 * 10^-^2$

$x^2+1.3*10^-^2x - 0.14$ × $1.3$ × $10^-^2= 0$

Solving the quadratic equation for $x , x \geq 0$ since $x$ represents a concentration;

$x=0.0366538$

Then, round the results to 2 significant figure;

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

Learn more about concentration here:

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3 0

2 years ago

Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of –7.2°C. Bromine can be classified as a

andreev551 [17]

Answer:

Pure substance.

Explanation:

7 0

3 years ago

You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?

umka21 [38]

Answer:

24.525 g of sulfuric acid.

Explanation:

Hello,

Normality (units of eq/L) is defined as:

$N=\frac{eq_{solute}}{V_{solution}}$

Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:

$eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq$

Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:

$m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4$

Best regards.

4 0

3 years ago

What noble gas configuration does oxygen achieve when bonded?

Greeley [361]

Answer:

1s2 2s2 2p6

Explanation:

ththe noble gas config. of neon

5 0

2 years ago

The charge of single electron

irga5000 [103]

The answer is B.

plz mark me as brainliest. i really need it.

8 0

3 years ago

When a deer eats too much in a given area, we call this _________.