# Page 24

Analysis: KHC<sub>8</sub>H<sub>4</sub>O<sub>4</sub>+NaOH-&gt;NaKC<sub>8</sub>H<sub>4</sub>O<sub>4</sub>+H<sub>2</sub>O

Trial 1:

10.00mL-0.20mL=<span style="text-decoration: underline;">9.80mL NaOH used</span>

2.00g/(39.10+1.008+8\*12.01+4\*1.008+4\*10.00)\*(1/1)\*(9.80/1000)=0.999~<span style="text-decoration: underline;">1.0M</span>

Trial 2:

19.70mL-10.00mL=<span style="text-decoration: underline;">9.70mL NaOH used</span>

2.00g/(39.10+1.008+8\*12.01+4\*1.008+4\*10.00)\*(1/1)\*(9.70/1000)=<span style="text-decoration: underline;">1.01M</span>

Trial 3:

29.40mL-19.70mL=<span style="text-decoration: underline;">9.7mL NaOH used</span>

2.00g/(39.10+1.008+8\*12.01+4\*1.008+4\*10.00)\*(1/1)\*(9.70/1000)=<span style="text-decoration: underline;">1.01M</span>

Trial 4:

39.15mL-29.40mL=<span style="text-decoration: underline;">9.75mL NaOH used</span>

2.00g/(39.10+1.008+8\*12.01+4\*1.008+4\*10.00)\*(1/1)\*(9.75/1000)=<span style="text-decoration: underline;">1.00M</span>

Trial 5:

48.85mL-39.15mL=<span style="text-decoration: underline;">9.70mL NaOH used</span>

2.00g/(39.10+1.008+8\*12.01+4\*1.008+4\*10.00)\*(1/1)\*(9.70/1000)=<span style="text-decoration: underline;">1.01M</span>

Average Molarity:

(1.00M+1.01M+1.01M+1.00M+1.01M)/5=1.006~<span style="text-decoration: underline;">1.01M on average</span>

% Deviation:

(1.01-1.00)/1.01 \*100=<span style="text-decoration: underline;">0.990% deviation</span>

Conclusion:

The purpose of the lab was to determine the molarity of an NaOH solution by reacting KHP with NaOH and getting as close to the endpoint as possible. We achieved this purpose by takings the before and after volume of the NaOH to when the enpoint is reached or passed, the difference of which was used to calculate the molarity of the NaOH solution. On average, 9.7mL was too far, though 9.75mL was used to get a near perfect titration. We got an average molarity of 1.01M NaOH. The target was about 1.00M.