Work the following problems on your own paper. Please staple your pages together

Question

Work the following problems on your own paper. Please staple your pages together before handing the problem set in. Each peoblem is worth 5 points, for a maximum of 35 points. Each problem is graded holistically on a scale of 0 - 5 based on accuracy and completeness. 1. (a) Calculate the pH and percent ionization for a 0.250 M solution of propionic acid (C_2H_5COOH), a monoprotic weak acid with K_0 = 1.35 times 10^-5. (b) Calculate the pH and percent ionization for a 0.250 M solution of propionic acid that also contains 0.100 M sodium propionate (NaC_2H_5COO). (c) Does the percent dissociation of C_2H_5COOH increase or decrease in the presence of the NaC_2H_5COO? Is this what you would expect based on LeChatelier's principle? Explain your answer! 2. (a) Calculate the pH of a buffer solution prepared by dissolving 5.23 grams of benzoic acid, C_6H_5COOH (FW = 122; K_a = 6.3times10^-5) and 4.47 grams of sodium benzoate (FW = 144), NaC_6H_5COO, into enough water to give 0.500 L of solution. (b) Calculate the pH of this buffer if the volume is increased to 2.00 L. 3. (a) Calculate the pH of the solution prepared by adding 2.50 times 10^-3 moles of NaOH to the buffer in 2a. Calculate the change in pH that occurs in the buffer upon adding the NaOH. (b) Calculate the pH of the solution prepared by adding 2.50 times 10^-3 moles of HCI to the butter in 2a. Calculate the change in pH that occurs in the buffer upon adding the HC1. 4. What is the pH of the solution formed by adding 25.0 mL of 0.150 M NaOH to 10.0 mL of0.240 M HCI? What is the pH of a solution formed by adding 25.0 mL of 0.150 M NaOH to 20.0 mL of 0.240 M HCIO_4? What is the pH of a solution formed by adding 25.0 mL of 0.150 M Ba(OH)_2 to 20.0 mL of 0.240 M HCIO_4? What is the pH of the solution prepared by adding 25.0 mL of 0.150 M KOH to 20.0 mL of 0.240 M HF (K_a = 6.8times10^-4)?

Explanation / Answer

4.-

NaOHpH=-Log(0.025*0.150)=2.426...14-2.426=11.574

HCLpH=-Log(0.010*0.240)=2.62...

(11.574+2.62)/2=7.097

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