Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:

Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
![\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}](https://tex.z-dn.net/?f=%5CKa%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BHCOO%5E-%5D%7D%7B%5BHCOOH%5D%7D)

From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,

Let's solve this for x. Multiply both sides by 0.12

taking square root to both sides:

Now, we have got the concentration of ![[H^+] .](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20.)
![[H^+] = 0.00465 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%200.00465%20M)
We know that, ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
Answer:
I'm sorry to say but there is no statements in this question meaning no one will be able to answer them with out them.. what are the statements?
Explanation:
Answer:
Pb²⁺(aq) --> Ni²⁺(aq)
Explanation:
The equation of the reaction is given as;
Ni(s) + Pb(NO₃)₂(aq) --> Pb(s) + Ni(NO₃)₂(aq)
In writing the ionic equation, we break the aqueous compound into ions. The solid and liquid compounds are ignored.
We have;
Pb²⁺(aq) + NO₃²⁻(aq) --> Ni²⁺(aq) + NO₃²⁻(aq)
Canceling the spectator ions;
Pb²⁺(aq) --> Ni²⁺(aq)
What the first person said lol