# A study of pH - lab related

1. 10.00 mL of 0.1000 M NH3 was diluted with distilled water to

40.00 mL. Calculate the pH of the dilute solution. (Kb NH3 = 1.8*10^-5)

2. The dilute solution was titrated with 0.0500 M HCl. Write the equation for the reaction between ammonia and hydrochloric acid. Calculate the pH of the solution after adding 4.00 mL of HCl.

3. Calculate the total volume of the solution at equivalence point.

4. Calculate the pH of the solution at equivalence point.

5. Calculate the pH of the solution after adding two additional drops of HCl after reaching the equivalence point (one drop ~ 0.05 mL)

(SHOW ALL CALCULATIONS, AS CLEARLY AS POSSIBLE)

https://brainmass.com/chemistry/acids-and-bases/study-ph-lab-related-16421

#### Solution Preview

1. The easiest way to solve this problem is by using the following formula:

M1V1=M2V2 (where M1 is molarity before dilution, V1 is the intial volume, M2 is molarity after dilution and V2 is the total final volume) and so 10x0.1=40xM2 and so M2=0.025M

this is our new NH3 concentration

we know that in H2O NH3 will undergo the following equilibrium:

NH3 + H2O --> NH4+ + OH- and that Kb= [OH][NH4+]/[NH3]

we will consider that at equilibrium x molar wil change from NH3 to products. and accoring to the mole rule (stiochiometry we know from the coefficients in the equation that 1 mole of NH3 gives 1 mole of NH4+ and 1 mole of OH- and so for x mole of NH3 undergoing acid/base change x moles of each NH4+ and OH- are generated)

and so 1.8X10^-5=x*x/(0.025-x)

for all puposes we can consider that x is much smaller than [NH3] and so 0.025-x=0.025

hence: 1.8*10^-5=x^2/0.025 and so x=6.7*10^-4

the [OH-]=6.7x10^-4 and so pOH= ...

#### Solution Summary

The solutions are given with short explanations.