### Introduction

** **Acids have a pH between 0 and 7, while bases have a pH between 7 and 14. A solution with a pH of 7 is said to be neutral; it is neither an acid nor a base. When given an acidic solution, it is possible add basic solution in order to neutralize it. To tell if the solution has been neutralized, an indicator such as phenolphthalein is used. The indicator will make the solution change color when it has become basic. In this experiment, a 10% solution of unknown molarity HCl was titrated with 1.00 * 10^{-1} M NaOH in order to neutralize the HCl. The molarity of the HCl was then able to be calculated knowing the molarity of NaOH, the volume of NaOH used, and the volume of HCl.

### Experimental

First, 25.00 ml of unknown sample of acid was delivered to a 250 mL volumetric flask using a volumetric pipette. The volumetric flask was then filled with de-ionized water to the mark in order to complete the 10% solution. A buret was then filled with 1.00 * 10^{-1} M NaOH solution and its starting point was recorded. Next, 25.00 mL of the acid solution was delivered to 300 mL Erlenmeyer flask and 3 drops of phenolphthalein were added. The Erlenmeyer flask was put under the buret and the NaOH solution was dispensed into the Erlenmeye flask until the indicator turned the solution pink for around 15 seconds. The final recording on the buret was recorded, and the process was performed 3 times in order to reduce error.

### Results

Molarity of NaOH solution: 1.00 * 10^{-1} M

Identification of acid: I

Trial 1 | Trial 2 | Trial 3 | |

Final Buret Reading, mL | 26.75 mL | 25.30 mL | 25.85 mL |

Initial Buret Reading, mL | 0.99 mL | 0.39 mL | 0.95 mL |

Volume of Titrant used, mL | 25.76 mL | 24.91 mL | 24.90 mL |

Volume of Titrant used, L | 0.02576 L | 0.02491 L | 0.02490 L |

Molarity of Diluted acid solution, M | 1.03 * 10^{-1} M | 9.96 * 10^{-2} M | 9.96 * 10^{-2} M |

Molarity of Undiluted acid solution, M | 1.03 M | 9.96 * 10^{-1} M | 9.96 * 10^{-1} M |

Mean Molarity of Undiluted acid solution, M: 1.01 M

### Calculations

** **In order to find the volume of titrant used in mL, I simply subtracted the initial buret reading from the final buret reading. To convert that volume in mL to L, I divided by 1000, as there are 1000 mL in 1 L. To find the molarity of diluted acid solution, I used the equation M_{1}V_{1} = M_{2}V_{2}. M_{1} equals the molarity of the NaOH solution used (1.00 * 10^{-1} M), V_{1 }equals the volume of titrant used in L, V_{2} equals the volume of acid solution used (0.02500 L), and M_{2} is the molarity of the acid solution, which was solved for. To find the molarity of the undiluted acid solution, I knew that a 10% solution was used, so I multiplied the molarity of the diluted solution by 10 to get the molarity of the undiluted solution. Lastly, to find the mean molarity of undiluted solution, I added the molarity of undiluted acid solution from the 3 trials and then divided by 3.

### Discussion/Conclusions

The final result of the molarity of the undiluted acid came very close to a whole number, which should mean my results are valid. It is also a very plausible number for the molarity of a solution. The volume of titrant used in my last two trials is nearly exact, so I must have performed them very well. The first trial is probably slightly off, as it was my first time doing a titration. On my first trial I added drops of NaOH too quickly towards the end. It is necessary to add the final drops and half drops very carefully towards the end, as it is a fine line between neutralizing the solution and making it basic. Overall, it was a very tedious experiment that relied on precision in order to achieve viable results.