JUST SHOW ME HOW TO DO at LEAST TWO .... a 20.00 mL sample of .200 M HBr ( a str

Question

JUST SHOW ME HOW TO DO at LEAST TWO .... a 20.00 mL sample of .200 M HBr ( a strong acid) is titratedwith .200 M NaOH solution. Calculate the pH of the solution afterthe following volumes of base have been added
a. 5. mL; b. 15.00 mL; c. 19.9 mL; d. 20.0 mL; e.20.1 mL; f. 35.0 mL; JUST SHOW ME HOW TO DO at LEAST TWO JUST SHOW ME HOW TO DO at LEAST TWO .... JUST SHOW ME HOW TO DO at LEAST TWO a 20.00 mL sample of .200 M HBr ( a strong acid) is titratedwith .200 M NaOH solution. Calculate the pH of the solution afterthe following volumes of base have been added
a. 5. mL; b. 15.00 mL; c. 19.9 mL; d. 20.0 mL; e.20.1 mL; f. 35.0 mL; JUST SHOW ME HOW TO DO at LEAST TWO

Explanation / Answer

Here is the reaction HBr + NaOH -> H2O + NaBr As you can see, this reaction is monoprotic acid titration;thus, the base and acid react at one to one ratio. In order to calculate the pH at each point, you just have tocalculate concentration of the H+ at each point. When 5 mL is titrated, # moles of H+ = (0.020 L HBr) * (0.200 HBr) - (0.005 L NaOH) *(0.200 M NaOH) = 0.003 moles H+ This is the number of moles of H+ after 5 mL of titration.Number of moles of H+ in the solution is given by = Number of molesof H+ present in the before titration - Number of moles NaOH istitrated. From here you just calculate the concentration of thesolution. Conc. of H+ at 5 mL = # of moles of H+/Total volume of thesolution = (0.003 moles H+) / (0.02+0.005)L = 0.12M Then just calculate the pH. pH = -log[H+] = -log[0.12] = 0.92 When 10 mL is titrated, # moles of H+ = (0.020 L HBr) * (0.200 HBr) - (0.015 L NaOH) *(0.200 M NaOH) = 0.001 moles H+ Conc. of H+ at 15 mL = 0.001moles H+/(0.02 + 0.015)L =0.2857 pH = -log[0.2857] = 1.54

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