Explain aqueous solution that has .025 moles of formic acid

Reference no: EM13490877

A beaker contains 250 mL of an aqueous solution that has .025 moles of formic acid (HCOOH) and .025 moles of sodium formate (HCOONa). The Ka of formic acid is 1.8 x 10^-4. How many moles of NaOH should be added to adjust the pH to 4.10. You can assume that the addition of NaOH does not change the volume of the solution. So here is what I have... 4.10= -log(1.8x10^-4) + log [base/acid] 4.10= -log(1.8^10-4) + log [.025+x/.025-x] .355= log [.025+x/.025-x] Now I have no way of getting to the next step. Anyone?

Reference no: EM13490877

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