Your post lists 0.086\it M as the concentration . Is that 0.086 M or jus what? I'll assume it is and you can adjust the answer below if it is something other than that.
Call the new acid HX, then
HX ==> H^+ + X^-
Write the Ka expression.
Ka = (H^+)(X^-)/(HX)
You know pH which you can convert to (H^+). (X^-) must be the same. (HX) = 0.086-(H^+). Solve for Ka and convert to pKa. Post your work if you get stuck.
A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's K_a value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries.
To find the pKa of X-281, you prepare a 0.086 \it M test solution of X-281. The pH of the solution is determined to be 2.70. What is the pKa of X-281?
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2 answers
4.32
4 answers