Calculate pH after adding 0.2mL of 1M HCl to 10mL of 1.0x10^-4M HCl

Question

pH?  

Answer 1

Hello Chris

 

Since HCl is a strong acid, it will ionize completely in an aqueous solution to form hydronium ions (H₃O⁺).

 

HCl + H₂O —> H₃O⁺ + Cl^-

 

From the given volumes and molarities, you want to first calculate the # of moles of HCl using the equation

 

mol = Molarity x Liter

 

For the 1.0 x 10^-4 M HCl solution,

 

mol = 1.0 x 10^-4 mol/L x 0.01 L = 1.0 x 10^-6 mol

 

For the 1M HCl solution,

mol = 1 mol/L x 0.0002 L= 2 X 10^-4 mol

 

Since 1M HCl is being added to the 1.0 x 10^-4 M solution, the moles of HCl calculated above become additive.

 

Therefore, the # of moles of HCl after the addition = 1.0 x 10^-6 + 2 x 10^-4 = 2.01 x 10^-4 mol ~ 2 x 10^-4 mol (rounded to the correct # of significant figures).

 

Similarly, the volumes of the 2 HCl solutions must also be added to get the final volume of the solution.

 

V_HCl = 0.01 L + 0.0002 L = 0.0102 L

 

From the total moles of HCl after the addition (2 x 10^-4 mol) and the total volume (0.0102 L), calculate the Molarity of the solution.

 

M = 2 x 10^-4 mol / 0.0102 L = 0.02 M (rounded to 1 significant figure).

 

The pH of this solution is then calculated using the equation,

 

pH = -log [H₃O⁺] = -log 0.02 = 1.7