## Chemistry and Chemical Reactivity (9th Edition)

[NaOH] = $0.167M$
This reaction takes place at the equivalence point; this means that that moles of KHC$_{8}$H$_{4}$O$_{8}$ = moles of NaOH Therefore to get [NaOH] we need the moles of NaOH used $\frac{0.902g KHC_{8}H_{4}O_{8}}{1}\times\frac{1 molKHC_{8}H_{4}O_{8} }{204.1gKHC_{8}H_{4}O_{8}}\times\frac{1 mol NaOH}{1 mol KHC_{8}H_{4}O_{8}} = 0.00442 mol NaOH$ To get [NaOH] divide moles by the concentration [NaOH] = $\frac{0.00442 mol NaOH}{0.02645 L} = 0.167M$