Suppose 25.0 mL of a solution containing 0.100 M H3PO4 is titrated with 0.200 M NaOH. Calculate...


Suppose 25.0 mL of a solution containing 0.100 M {eq}H_3PO_4 {/eq} is titrated with 0.200 M {eq}NaOH {/eq}. Calculate the pH of the solution after 20.0 mL of {eq}NaOH {/eq} has been added.


Titration is an analytical technique that is used in the determination of the concentration of an unknown compound. In an acid-base titration, a direct transfer of protons from acid to the hydroxide is involved. Titration can be done in a weak acid and strong base, or a weak base and a strong acid.

Answer and Explanation: 1

The balanced chemical reaction of the problem is shown below:

{eq}\rm H_3PO_4 + OH^- \rightarrow H_2PO_4^- + H_2O {/eq}

The total volume of the solution is

{eq}\rm Total\,Volume = 25.0\,mL + 20\,mL = 45\,mL =0.0450\,L {/eq}

Solving for the concentration of each component:

{eq}\rm [H_3PO_4]=\dfrac{(0.0250\,L)(0.100\,M)}{(0.0450\,L)}=0.0556\,M\\ [OH^-]=\dfrac{(0.0200\,L)(0.200\,M)}{0.0450\,L)}=0.0889\,M {/eq}

Constructing the ICE table:

{eq}\rm Total\,Volume = 25.0\,mL + 20\,mL = 45\,mL =0.0450\,L {/eq}


Since sodium hydroxide is a strong base and is present in a larger quantity, the phosphoric acid initially present will be consumed to neutralize the added base. Hence, we can now solve for the pH by solving the pOH first:

{eq}\rm pOH=-log([OH^-])\\ pOH=-log(0.0333)\\ pOH=1.48 {/eq}

Solving for the pH of the solution:

{eq}\rm pH=14-pOH\\ pH=14-1.48\\ \boxed{\mathbf{pH=12.5}} {/eq}

Learn more about this topic:

Titrations with Weak Acids or Weak Bases


Chapter 11 / Lesson 10

Learn about titrations of weak acids and strong bases, strong acids and weak bases, and weak acids and weak bases. Explore types of titration curves and the pH.

Related to this Question

Explore our homework questions and answers library