I'm not really sure how to go about doing this problem...

What is the pH of a 7.40 x 10-3 M solution of NH4Cl?
Help is appreciated! Thank you

To find the pH of the solution, we need to understand the concept of pH and how to calculate it.

The pH scale is a measure of how acidic or basic a solution is. It ranges from 0 to 14, with values less than 7 considered acidic, values greater than 7 considered basic, and a pH of 7 considered neutral.

In this case, we have a solution of NH4Cl. NH4Cl is the salt of a weak base (NH3) and a strong acid (HCl). It dissociates in water, producing NH4+ and Cl- ions.

To calculate the pH of this solution, we need to consider the dissociation of NH4+ in water. NH4+ can react with water and donate a proton (H+), making the solution acidic.

The dissociation reaction is as follows:
NH4+ + H2O ⇌ NH3 + H3O+

The concentration of NH4Cl is given as 7.40 x 10-3 M, but it assumes complete dissociation, resulting in a concentration of NH4+ ions equal to 7.40 x 10-3 M.

Now, we can proceed to calculate the pH using the concept of pOH. pOH is the negative logarithm (base 10) of the hydroxide ion concentration (OH-) in a solution.

Since we are dealing with NH4+, we need to find the concentration of the hydroxide ions (OH-) using the autoionization of water. The autoionization of water is as follows:
H2O ⇌ H+ + OH-

The concentration of H+ is equal to the concentration of OH- in pure water, which is 1 x 10-7 M (at 25 degrees Celsius).

However, in this case, we have a weak acid (NH4+) that will release H+ ions, making the concentration of H+ ions higher than 1 x 10-7 M.

To calculate the concentration of OH- ions, we can use the equation Kw = [H+][OH-]. Kw is the ion product of water and is equal to 1 x 10^-14 (at 25 degrees Celsius).

So, [H+][OH-] = 1 x 10^-14.
If we assume that [H+] ≈ [OH-], then we can substitute [OH-] with [H+] in the equation above.

Thus, [H+][H+] = 1 x 10^-14.
Simplifying, we get [H+]^2 = 1 x 10^-14.

Taking the square root of both sides, we have [H+] = √(1 x 10^-14) = 1 x 10^-7 M.

Now, we have the concentration of H+ ions. To calculate the pOH, we can use the equation pOH = -log[OH-].

Since [OH-] = [H+], we can rewrite the equation as pOH = -log[H+].

Thus, pOH = -log(1 x 10^-7) = 7.

Finally, to calculate the pH, we can use the equation pH + pOH = 14. Rearranging the equation, we get pH = 14 - pOH.

Hence, pH = 14 - 7 = 7.

Therefore, the pH of the 7.40 x 10-3 M solution of NH4Cl is 7.