Question

Determine the pH of a solution when 21.8 mL of 0.026 M HNO3 is mixed with 17.4 mL of A) 0.0100 M NaOH.

Find the pH:

a) 0.0100 M NaOH

b.) distilled water

c.) 0.0060 M HCl

d.) 0.120 M KOH

Answer 1

both HNO3 and NaOH are strong acid and base respectively.

moles = molarity* Volume in L

moles : HNO3= 0.026*21.8/1000 =0.000567 NaOH= 0.01*17.4/1000 =0.000174

the reaction between HNO3 and KOH is HNO3+ NaOH------> NaNO3+ H2O

theoretical molar ratio of HNO3: KOH= 1:1, actual molar ratio = 0.000567:0.000174= 3.25:1

excess is HNO3. moles of HNO3 excess= 0.000567-0.000174 =0.0003928 the remaining HNO3 and NaOH undergo reaction and get neutralized.

volume of mixing = 21.8+17.4= 39.2ml= 39.2/1000 L= 0.0392L

concentration of HNO3=0.0003928/0.0392 =0.01M

HNO3 ionizes completely as HNO3+ H2O------>H3O++ NO3-

pH= -log[H3O+]= -log (0.01)= 2

2. HCl is also strong acid . hence [H3O+] from HCl =0.006, [H3O+] from HNO3=0.026

total of [H+] = 0.006+0.026 =0.032, pH= -log [H+]= -log (0.032)= 1.49

4. moles of KOH in 0.12M of 21.8 ml= 0.12*21.8/1000 =0.002616

the reaction between KOH and HNO3 is HNO3+ KOH----> KNO3+ H2O

molar ratio of HNO3: KOH =1:1, actual ratio =0.000567 :0.002616

so excess is KOH and all the HNO3 reacts. Remaining KOH=0.002616-0.000567= 0.002049

concentration of KOH= moles/ volume in L =0.002049*1000/21.8 =0.094M

KOH being strong base ionizes completely, [OH-]= 0.094

pOH= -log (0.094)=1.03

pH= 14-1.03 =12.97