A 30.00-mL sample of a weak acid is titrated with 0.0167 M NaOH. At the endpoint

Question

A 30.00-mL sample of a weak acid is titrated with 0.0167 M NaOH. At the endpoint, it is found that 40.35 mL of titrant was used. What was the concentration of the weak acid?

Explanation / Answer

First, find out how many moles of NaOH was used by understanding that 0.0167 M means 0.0167 moles per liter. 40.35mL of NaOH= .04035 L of NaOH If we multiple molarity(moles/liter) by liters, the liters will cancel out, leaving moles. .04035 L of NaOH x 0.0167 M NaOH = 6.738 x 10^-4 moles of NaOH. Since we know that the net ion eqution is H(+) + OH(-) ---> H2O We know for every one mole of NaOH(which is the same amount of just OH in this case), one mole of the weak acid will be used. Since we had to use 6.738 x 10^-4 moles of NaOH, we had 6.738 x 10^-4 moles of the acid. From here, we just convert milliliters to liters and divide our amount of moles(6.738 x 10^-4) by the liters. 30ml = .03 L 6.738 x 10^-4 moles / .03 L = .02246 M of the weel acid solution. So, after rounding, the answer is E, .0225 M.

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