Answer:
Ka = 0.1815
Explanation:
Chromic acid
pH = ?
Concentration = 0.078 M
Ka = ?
HCl
conc. = 0.059M
pH = -log(H+)
pH = -log(0.059) = 1.23
pH of chromic acid = 1.23
Step 1 - Set up Initial, Change, Equilibrium table;
H2CrO4 ⇄ H+ + HCrO4−
Initial - 0.078M 0 0
Change : -x +x +x
Equilibrium : 0.078-x x x
Step 2- Write Ka as Ratio of Conjugate Base to Acid
The dissociation constant Ka is [H+] [HCrO4−] / [H2CrO4].
Step 3 - Plug in Values from the Table
Ka = x * x / 0.078-x
Step 4 - Note that x is Related to pH and Calculate Ka
[H+] = 10^-pH.
Since x = [H+] and you know the pH of the solution,
you can write x = 10^-1.23.
It is now possible to find a numerical value for Ka.
Ka = (10^-1.23))^2 / (0.078 - 10^-1.23) = 0.00347 / 0.0191156
Ka = 0.1815
In 1(atm) the patients blood is 760 mmHg
The answer is D) the bonding in sodium chloride crystals is strong
Answer:
Explanation:
In the qualitative analysis of metal salts , we see that in group I , metal chlorides are precipitated out . It is so because their metal chlorides are insoluble in water .
In this group following metal ions are present
Ag+,
Hg₂²⁺
Pb²⁺
