Question

Calculate the pH of the resulting solution if 29.0 mL of 0.290 M HCl(aq) is added to 39.0 mL of 0.290 M NaOH(aq).
pH=
Calculate the pH of the resulting solution if 29.0 mL of 0.290 M HCl(aq) is added to 19.0 mL of 0.390 M NaOH(aq).
pH=

Answer 1

We have the given reaction as;

---->  

Answer A) The pH will be 12.36,

We have to convert the concentrations of HCl and NaOH into moles,

So we have, n(HCl) = (0.0290 L) X (0.290 mol/L) = 8.41X  moles  

and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13X  moles  

Now, it seems NaOH is in excess, so amount remaining will be;  

1.13 X  - 8.41 X  = 2.89 X  moles  

Now, the total volume will become as  = 0.0390 + 0.0290 = 0.068 L  

So, the concentration of [] = 2.89×10ˉ³ mol / 0.068 L = 4.25 X  M  

pOH = - log [] = -log (4.25×) = 1.37  

Hence, pH = 14 - pOH = 14 - 1.37 = 12.6

So the pH of the solution will be 12.6 which is basic in nature.  

Answer B) The pH will be 1.68  

Now, for the given concentration we need to find moles for HCl and NaOH also;

n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41 X  mol  

n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X   mol  

here we can see, HCl is in excess amount so the remaining will be;  

8.41X  - 7.41 X  = 1.0 X  mol  

Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L  

So the concentration of [HCl] = 1.0 X  mol / 0.0480 L = 2.08 X  M

 

Which is = [H⁺]  

So, the pH = - log [] = -log(2.08X ) = 1.68

 

Hence, the pH will be 1.63 which is more acidic in nature.