A titration of 65.0 mL of a tartaric acid (H,C,HO.) solution of unknown concentration with 0.805 M LIOH requires 32.0 mL to reach the second equivalence point. What is the concentration of the tartaric acid solution?

Consider the titration of 80.0 mL of 0.100 M Ba (OH), by 0.400 M HCI. Calculate the pH of the resulting solution after the following volumes of HCl have been added. а. 0.0 mL pH = b. 20.0 mL pH = c. 30.0 mL pH = d. 40.0 ml pH = e. 90.0 ml pH =

When pH titrations occur, more data points become necessary at the vertical deflection near the equivalence point which is accomplished by adding smaller volumes of base from the pipers before taking the pH of the solution. Why are more points needed at this point in the titration curve? Why can larger volumes be used in the area outside of this region?

Which is the best and easiest way to do this problem ?

1) If you weigh out 0.2349 g of tris, what volume of the 0.1 M HCI solution will you need to reach the end point of this titration? (Note: this is a nice way to get the approximate volume before starting a titration!) 2) If you weighed out 0.4670 g of KHP, what volume of the 0.1 M NaOH solution will you need to reach the end point of this titration?

O ACIDS AND BASES Calculating the pH of a weak base titrated with a strong acid with a 0.3000 M solution of HNO,. The p K, of An analytical chemist is titrating 160.2 mL of a 0.5000 M solution of dimethylamine ((CH₂)2NH) w dimethylamine is 3.27. Calculate the pH of the base solution after the chemist has added 280.8 mL of the HNO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added. Round your answer to 2 decimal places... PH-0 X

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