A volume of 500.0 mL of 0.180 M NaOH is added to 595 mL of 0.200 M weak acid ($K_a = 1.99 \times 10^{-5}$). What is the pH of the resulting buffer? HA(aq) + OH$^-$(aq) $\rightarrow$ H$_2$O(l) + A$^-$(aq) 1.38077 pH = Incorrect
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Moles of NaOH = (0.500 L)(0.180 mol/L) = 0.090 mol Moles of weak acid = (0.595 L)(0.200 mol/L) = 0.119 mol Show more…
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