How many mL of concentrated H2SO4 are needed to decrease the pH to 1.5 of 10 mL of a 0.1 M NaOH solution?

Question

How many mL of concentrated H2SO4 are needed to decrease the pH to 1.5 of 10 mL of a 0.1 M NaOH solution?

 

Thanks

Answer 1

So, this is, or can be, a rather complicated calculation depending on the assumptions, etc.  However, in its simplest form..

 

2NaOH + H2SO4 ==> Na2SO4 + 2H2O

Assuming a concentration of H2SO4 of 18.4 M (typical for concentrated H2SO4), first determine volume of H2SO4 needed to completely neutralize 10 ml of 0.1 M NaOH:

• 0.010 L x 0.1 mol/L = 1x10^-3 moles NaOH present
• 1x10^-3 moles NaOH x 1 mole H2SO4/2 moles NaOH = 5x10^-4 moles H2SO4 for complete neutralization (pH = 7)
• (x L)(18.4 mol/L) = 5x10^-4 moles and x = 2.7x10^-5 L of H2SO4 needed for complete neutralization (ph = 7)

 

Next, determine volume H2SO4 to go from pH 7 to pH 1.5

• Assume that the volume (2.7x10^-5 L or 27 ul) did not change the volume of the 10 mls so you still have 10 ml
• pH 1.5 means [H⁺] = 3.2x10^-2 M
  
• (x L)(18.4 M) = (0.010 L)(3.2x10^-2 M) and x = 1.7x10^-5 L (17 ul) H2SO4 needed.  This also assumes that only the first ionization of H2SO4 is considered in supplying the H⁺

 

Thus, the total volume of 18.4 M H2SO4 needed to decrease the pH of 10 ml of 0.1 M NaOH to 1.5 would be 27 ul + 17 ul = 44 ul = 0.044 mls