# What is acidic neutralization

Example \(\PageIndex{1}\): Titrating a Weak Acid

Suppose 13.00 mL of a weak acid, with a molarity of 0.1 M, is titrated with 0.1 M NaOH. How would we draw this titration curve?

**SOLUTION**

**Step 1: **First, we need to find out where our titration curve begins. To do this, we find the initial pH of the weak acid in the beaker before any NaOH is added. This is the point where our titration curve will start. To find the initial pH, we first need the concentration of H_{3}O^{+}.

Set up an ICE table to find the concentration of H3O+:

\(HX\) | \(H_2O\) | \(H_3O^+\) | \(X^-\) |
---|---|---|---|

Initial | 0.1M | ||

Change | -xM | +xM | +xM |

Equilibrium | (0.1-x)M | +xM | +xM |

\[Ka=(7)(10^{-3})\]

\[K_a=(7)(10^{-3})=\dfrac{(x^2)M}{(0.1-x)M}\]

\[x=[H_3O^+]=0.023\;M\]

Solve for pH:

\[pH=-\log_{10}[H_3O^+]=-\log_{10}(0.023)=1.64\]

**Step 2: **To accurately draw our titration curve, we need to calculate a data point between the starting point and the equivalence point. To do this, we solve for the pH when neutralization is 50% complete.

Solve for the moles of OH- that is added to the beaker. We can to do by first finding the volume of OH- added to the acid at half-neutralization. 50% of 13 mL= 6.5mL

Use the volume and molarity to solve for moles (6.5 mL)(0.1M)= 0.65 mmol OH^{-}

Now, Solve for the moles of acid to be neutralized (10 mL)(0.1M)= 1 mmol HX

Set up an ICE table to determine the equilibrium concentrations of HX and X:

\(HX\) | \(H_2O\) | \(H_3O^+\) | \(X^-\) |
---|---|---|---|

Initial | 1 mmol | ||

Added Base | 0.65 mmol | ||

Change | -0.65 mmol | -0.65 mmol | -0.65 mmol |

Equilibrium | 0.65 mmol | 0.65 mmol |

To calculate the pH at 50% neutralization, use the Henderson-Hasselbalch approximation.

pH=pKa+log[mmol Base/mmol Acid]

pH=pKa+ log[0.65mmol/0.65mmol]

pH=pKa+log(1)

\[pH=pKa\]

Therefore, when the weak acid is 50% neutralized, pH=pKa

**Step 3: **Solve for the pH at the equivalence point.

The concentration of the weak acid is half of its original concentration when neutralization is complete 0.1M/2=.05M HX

Set up an ICE table to determine the concentration of OH-:

\(HX\) | \(H_2O\) | \(H_3O^+\) | \(X^-\) |
---|---|---|---|

Initial | 0.05 M | ||

Change | -x M | +x M | +x M |

Equilibrium | 0.05-x M | +x M | +x M |

Kb=(x^2)M/(0.05-x)M

Since Kw=(Ka)(Kb), we can substitute Kw/Ka in place of Kb to get Kw/Ka=(x^2)/(.05)

\[x=[OH^-]=(2.67)(10^{-7})\]

\[pOH=-\log_{10}((2.67)(10^{-7}))=6.57\]

\[pH=14-6.57=7.43\]

**Step 4: **Solve for the pH after a bit more NaOH is added past the equivalence point. This will give us an accurate idea of where the pH levels off at the endpoint. The equivalence point is when 13 mL of NaOH is added to the weak acid. Let's find the pH after 14 mL is added.

Solve for the moles of OH-

\[ (14 mL)(0.1M)=1.4\; mmol OH^-\]

Solve for the moles of acid

\[(10\; mL)(0.1\;M)= 1\;mmol \;HX\]

Set up an ICE table to determine the \(OH^-\) concentration:

\(HX\) | \(H_2O\) | \(H_3O^+\) | \(X^-\) |
---|---|---|---|

Initial | 1 mmol | ||

Added Base | 1.4 mmol | ||

Change | -1 mmol | -1 mmol | 1 mmol |

Equilibrium | 0 mmol | 0.4 mmol | 1 mmol |

\[[OH-]=\frac{0.4\;mmol}{10\;mL+14\;mL}=0.17\;M\]

\[pOH=-log_{10}(0.17)=1.8\]

\[pH=14-1.8=12.2\]

We have now gathered sufficient information to construct our titration curve.

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