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

## Question:

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:

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}

initial | 0.0556 | 0.0889 | 0 |

change | -0.0556 | -0.0556 | +0.0556 |

equilibrium | 0 | 0.0333 | 0.0556 |

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:

from

Chapter 11 / Lesson 10Learn 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.