For the equilibrium that exists in an aqueous solution<span> of nitrous acid (</span>HNO2, a weak acid) ... [H+][NO2. –]. [HNO2<span>]. PAGE: 14.1. 2. Which of the following is a conjugate ... Using the following Ka values, indicate the correct </span>order<span> of base strength. </span>HNO2<span>. Ka = </span>4.0<span> × 10–4 .... Calculate the [H+] in a </span>solution<span>that has a </span>pH<span> of 11.70.
i hope thid works
</span>
Answer:
0.3023 M
Explanation:
Let Picric acid = 
So, + 
⇄ 
+ 
The ICE table can be given as:
+ ⇄ +
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
= 
OR 
= 
OR 
= 
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [ ] = [ ] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).
The pH of a neutral aqueous solution at 37°C is 6.8.
<h3>What is Kw? </h3>
Kw is defined as the dissociation, which is also known as self-ionization, constant of water. this is an equilibrium constant, and its expression is:
Kw = [OH⁻] . [H₃O⁺]
Neutral pH determines that the concentrations of OH⁻ and H₃O⁺ are equal.
<h3>Calculation</h3>
Let us suppose concentration of OH and H₃O⁺ is x, to calculate it:
Kw =[OH⁻] . [H₃O⁺] = x²
x² = 2.4 × 10⁻¹⁴ M²
x = 1.5919 × 10⁻⁷ M
Hence, the concentration of OH and H₃O⁺ (x) = [H₃O⁺] = [OH⁻] = 1.5919×10⁻⁷ M
pH = -log[H₃O⁺] = -log( 1.5919×10⁻⁷ M)
pH = 6.8
Thus, we find that the pH of a neutral aqueous solution at 37 °c (which is the normal human body temperature) is 6.8.
learn more about pH:
brainly.com/question/9529394
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