What is the pH of 40.0 mL of 0.100 M HAc mixed with 30.0 mL of 0.100 M NaOH and 10.0 mL of distilled H2O?
We know that, pH is the negative logarithm of hydrogen ion concentration. i.e.
pH = - log[H+]
Now, on adding NaOH solution to an acid solution, 1 : 1 molar neutralization reaction takes place, as follows,
HAc + NaOH -----> AcNa + H2O
Thus, 1 mole of HAc is neutralized by 1 mole of NaOH
Now, for a neutralization reaction,
moles of H+ neutralized = moles of OH- added
=> moles of HAc neutralized = moles of NaOH added
=> moles of HAc neutralized= Molarity of NaOH x Volume of NaOH added
= 0.100 M x 30.0 mL
= 0.100 mol/L x 0.030 L
= 0.003 mol
Thus, remaining moles of HAc in the solution = Initial moles of HAc - moles of HAc neutralized
= ( Molarity of HAc x Volume of HAc ) - moles of HAc neutralized
= ( 0.100 M x 40.0 mL) - 0.003 mol
= ( 0.100 mol/L x 0.040 L) - 0.003 mol
= 0.004 - 0.003 mol
= 0.001 mol
Now, total volume of solution = 40 mL + 30 mL + 10 mL
= 80 mL
= 0.080 L
So, remaining HAc concentration in the solution is,
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