What is the ph of a solution prepared by mixing 50.00 ml of 0.10 m nh₃ with 25.00 ml of 0.10 m nh₄cl?
assume that the volume of the solutions are additive and that kb = 1.8 × 10⁻⁵ for nh₃. enter your answer in exponential (e) format (sample 1.23 e-4) with two decimal places and without units.

9.56

Explanation:

First we need to know the moles of each species involved:

NH3: 50.00* 10^-3* 0.10 M =5.0* 10^-3

NH4Cl: 25.00* 10^-3* 0.10 M = 2.5 * 10^-3

Then we calculate the molarities:

5.0* 10^-3 moles/75 * 10^-3 L =0.67

2.5 * 10^-3 moles/75 * 10^-3 L =0.33

Ka and Kb are related by:

pKa = 14 - pKb

If Kb is 1.8 *10^-5

then pKb is - log (1.8 *10^-5)=4.74

Therefore

pKa =14 -4.74 =9.26

Using the Henderson-Hasselbalch (H-H) equation,

pH = pKa + log (0.67)/(0.33) = 9.26+ log 2 =9.26 + 0.30 = 9.56

idk

explanation: