Question

Estimate the volume of a solution of 4.5 M NaOH that must be added to adjust the pH from 4.0 to 9.0 in 78 ml of a 81 mM solution of phosphoric acid. ml

Answer 1

Given that, Molarity of Phosphoric acid = 81mM = 81/1000 M = 0.081M [ $\because$ 1M = 1000mM]

Volume of Phosphoric acid = 78 mL = 0.078 L . [ $\because$ 1 L = 1000mL]

Moles of Phosphoric acid = Volume * Molarity = 0.081 M * 0.078 L = 0.006318 moles

Now, the reaction between Phosphoric acid and NaOH is :

H₃PO₄ + NaOH   $\rightarrow$ NaH₂PO₄ + H₂O

Thus, 1 mol of Phosphoric acid will neutralise by 1 mol of NaOH.

Thus, moles of NaOH required =  0.006318 moles

Molarity of NaOH solution = 4.5 M = 4.5 mol/L

Therefore, volume of NaOH = Moles/ Molarity = 0.006318 moles/ (4.5 mol/L) = 0.00140 L

= 1.40 mL.

Thus, volume of NaOH that must be added = 1.40 mL