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.
Answer:
its B
Explanation:
Because its can only go cold to hot. -hope this help :)
The balanced equation for the reaction between Ba(OH)₂ and HCl is as follows;
Ba(OH)₂ + 2HCl ---> BaCl₂ + 2H₂O
stoichiometry of Ba(OH)₂ to HCl is 1:2
the number of HCl moles that have reacted - 0.2452 mol/L x (20.00 x 10⁻³ L)
number of HCl moles reacted = 0.004904 mol
2 mol of HCl reacts with 1 mol of Ba(OH)₂
therefore 0.004904 mol of HCl reacts with - 1/2 x 0.004904 mol of Ba(OH)₂
number of Ba(OH)₂ moles in 18.15 mL - 0.002452 mol
Therefore number of Ba(OH)₂ moles in 1000 mL- 0.002452 mol /(18.15 x 10⁻³ L)
molarity of Ba(OH)₂ is = 0.1351 M
