Answer:
HNO₂
Explanation:
An acid is a proton donor; a base is a proton acceptor.
Thus, NO₂⁻ is the base, because it accepts a proton from the water.
H₂O is the acid, because it donates a proton to the nitrite ion.
The conjugate base is what's left after the acid has given up its proton.
The conjugate acid is what's formed when the base has accepted a proton.
NO₂⁻/HNO₂ make one conjugate acid/base pair, and H₂O/OH⁻ are the other conjugate acid/base pair.
NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻
base acid conj. conj.
acid base
Answer:
Solution A is 1,000 times more acidic than Sol. B
Explanation:
for pH values we use scientific notation:
-log10 c (where c is the hydrogen ion concentration) is used to notate pH value (think of it as a unit)
ie:
10^-2 is sol A 10^-5 is sol B
5-2 is 3
10^-3 = 1000
there's a diff of 1,000 between the solutions.
The <span>era that is know as the age of mammals is the </span><span>Mesozoic
period.
Answer: Letter D
Hope that helps. -UF aka Nadoa</span>
Answer:
11.9 is the pOH of a 0.150 M solution of potassium nitrite.
Explanation:
Solution : Given,
Concentration (c) = 0.150 M
Acid dissociation constant = 
The equilibrium reaction for dissociation of 
(weak acid) is,
initially conc. c 0 0
At eqm. 
First we have to calculate the concentration of value of dissociation constant 
.
Formula used :
Now put all the given values in this formula ,we get the value of dissociation constant .
By solving the terms, we get
No we have to calculate the concentration of hydronium ion or hydrogen ion.
![[H^+]=c\alpha=0.150\times 0.0533=0.007995 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%3D0.150%5Ctimes%200.0533%3D0.007995%20M)
Now we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
pH + pOH = 14
pOH =14 -2.1 = 11.9
Therefore, the pOH of the solution is 11.9
Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
