Explanation:
The atomic number is equal to the number of protons in an atom's nucleus. Hydrogen's atomic number is 1 because all hydrogen atoms contain exactly one proton.
Answer:
Here's what I get
Explanation:
1. Nickel sulfate
base + acid ⟶ salt + water
NiSO₄ is a salt of the base Ni(OH)₂ and the acid sulfuric acid.
Hydroxides of transition metals are insoluble; most sulfates are soluble.
![\underbrace{\hbox{Ni(OH)$_{2}$(s)}}_{\hbox{base}} + \underbrace{\hbox{H$_{2}$SO$_{4}$(aq)}}_{\hbox{acid}} \longrightarrow \, \underbrace{\hbox{NiSO$_{4}$(aq)}}_{\hbox{salt}} + \underbrace{\hbox{2H$_{2}$O(l)}}_{\hbox{water}}](https://tex.z-dn.net/?f=%5Cunderbrace%7B%5Chbox%7BNi%28OH%29%24_%7B2%7D%24%28s%29%7D%7D_%7B%5Chbox%7Bbase%7D%7D%20%2B%20%5Cunderbrace%7B%5Chbox%7BH%24_%7B2%7D%24SO%24_%7B4%7D%24%28aq%29%7D%7D_%7B%5Chbox%7Bacid%7D%7D%20%5Clongrightarrow%20%5C%2C%20%5Cunderbrace%7B%5Chbox%7BNiSO%24_%7B4%7D%24%28aq%29%7D%7D_%7B%5Chbox%7Bsalt%7D%7D%20%2B%20%5Cunderbrace%7B%5Chbox%7B2H%24_%7B2%7D%24O%28l%29%7D%7D_%7B%5Chbox%7Bwater%7D%7D)
2. Carbonate + acid
Most carbonates are insoluble.
They react with acids to form carbonic acid (H₂CO₃), which decomposes into water and carbon dioxide.
![\rm NiCO_{3}(s) + H_{2}SO_{4}(aq) \longrightarrow \, NiSO_{4}(aq) + H_{2}O(l) + CO_{2}(g)](https://tex.z-dn.net/?f=%5Crm%20NiCO_%7B3%7D%28s%29%20%2B%20H_%7B2%7DSO_%7B4%7D%28aq%29%20%5Clongrightarrow%20%5C%2C%20NiSO_%7B4%7D%28aq%29%20%2B%20H_%7B2%7DO%28l%29%20%2B%20CO_%7B2%7D%28g%29)
Answer:
9.36
Explanation:
Sodium formate is the conjugate base of formic acid.
Also,
![K_a\times K_b=K_w](https://tex.z-dn.net/?f=K_a%5Ctimes%20K_b%3DK_w)
for sodium formate is ![K_b=\frac {K_w}{K_a}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%20%7BK_w%7D%7BK_a%7D)
Given that:
of formic acid = ![1.8\times 10^{-4}](https://tex.z-dn.net/?f=1.8%5Ctimes%2010%5E%7B-4%7D)
And, ![K_w=10^{-14}](https://tex.z-dn.net/?f=K_w%3D10%5E%7B-14%7D)
So,
![K_b=\frac {10^{-14}}{1.8\times 10^{-4}}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%20%7B10%5E%7B-14%7D%7D%7B1.8%5Ctimes%2010%5E%7B-4%7D%7D)
![K_b=5.5556\times 10^{-11}](https://tex.z-dn.net/?f=K_b%3D5.5556%5Ctimes%2010%5E%7B-11%7D)
Concentration = 0.35 M
HCOONa ⇒ Na⁺ + HCOO⁻
Consider the ICE take for the formate ion as:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻
At t=0 0.35 - -
At t =equilibrium (0.35-x) x x
The expression for dissociation constant of sodium formate is:
![K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5BOH%5E-%5D%5BHCOOH%5D%7D%7B%5BHCOO%5E-%5D%7D)
![5.5556\times 10^{-11}=\frac {x^2}{0.35-x}](https://tex.z-dn.net/?f=5.5556%5Ctimes%2010%5E%7B-11%7D%3D%5Cfrac%20%7Bx%5E2%7D%7B0.35-x%7D)
Solving for x, we get:
x = 0.44×10⁻⁵ M
pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64
pH + pOH = 14
So,
<u>pH = 14 - 4.64 = 9.36</u>