Answer:
The pH in 0.140 M hippuric acid solution is 2.2.
Explanation :
Dissociation constant of the acid = 
![pK_a=-\log[K_a]](https://tex.z-dn.net/?f=pK_a%3D-%5Clog%5BK_a%5D)
![3.62=-\log[K_a]](https://tex.z-dn.net/?f=3.62%3D-%5Clog%5BK_a%5D)
Concentration of hippuric acid = c = 0.140 M
Initially
c 0 0
At equilibrium
(c-x) x x
Concentration of acid = c ![[HC_9H_8NO_3]=0.140 M](https://tex.z-dn.net/?f=%20%5BHC_9H_8NO_3%5D%3D0.140%20M)
Dissociation constant of an acid is given by:
![K_a=\frac{[C_9H_8NO_{3}^-][H^+]}{[HC_9H_8NO_{3}]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BC_9H_8NO_%7B3%7D%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHC_9H_8NO_%7B3%7D%5D%7D)
Solving for x:
x = 0.005677 M
![[H^+]=x = 0.005677 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%20%3D%200.005677%20M)
The pH of the solution :
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![pH=-\log[0.005677 M]=2.246\approx 2.2](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B0.005677%20M%5D%3D2.246%5Capprox%202.2)
The pH in 0.140 M hippuric acid solution is 2.2.
What happens if you apply pressure on the oobleck?
the surfice gets hard and nothing can pass through it, you can punch it and it will hurt
What happens if you try rolling the oobleck into a ball?
you cant i will go back to being a liquid and fall through your fingers
What happens when you let it run through your fingers?
its just like a liquid
only if you add presure will it turn hard
Answer:
(a) Acid
(b) Base
(c) Acid
(d) Base
Explanation:
According to the Arrhenius acid-base theory:
- An acid is a substance that releases H⁺ in aqueous solution.
- A base is a substance that releases OH⁻ in aqueous solution.
(a) H₂SO₄ is an acid according to the following equation:
H₂SO₄(aq) ⇒ 2 H⁺(aq) + SO₄²⁻(aq)
(b) Sr(OH)₂ is a base according to the following equation:
Sr(OH)₂(aq) ⇄ Sr²⁺(aq) + 2 OH⁻(aq)
(c) HBr is an acid according to the following equation:
HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)
(d) NaOH is a base according to the following equation:
NaOH(aq) ⇒ Na⁺(aq) + OH⁻(aq)
<span>[H3O+] = 10^(-pH) = 10^(-4.20) = 6.3 x 10^-5 M
pOH = 14 - pH = 14 - 4.20 = 9.80
[OH-] = 10^(-pOH) = 10^(-9.80) = 1.6 x 10^-10 M
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