An aqueous solution at room temperature contains 0.1 M NH4Cl and 0.01 M NH4OH (pKb = 5), the pH of the solution is :-7.56.86.58.0

The problem is asking for the pH of a solution containing both a weak base (NH4OH) and its conjugate acid (NH4Cl). This is a buffer solution, and we can use the Henderson-Hasselbalch equation to solve for the pH.

The Henderson-Hasselbalch equation is: pH = pKa + log([A-]/[HA])

However, since we are dealing with a weak base and its conjugate acid, we need to use the form of the equation that deals with pOH and pKb, which is: pOH = pKb + log([HB+]/[B])

Given: pKb = 5 [HB+] = concentration of NH4Cl = 0.1 M [B] = concentration of NH4OH = 0.01 M

Substitute these values into the equation:

pOH = 5 + log(0.1/0.01)

Solving the equation gives:

pOH = 5 + log(10) = 5 + 1 = 6

Remember that pH + pOH = 14 at room temperature, so we can solve for pH:

pH = 14 - pOH = 14 - 6 = 8

So, the pH of the solution is 8.0.