Michael P. answered • 02/27/18

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L water = 2.5 fl oz x 1 gal/128 fl oz x 3.78541 L/gal = 0.07393 L water

x = L H

_{2}SO_{4 }needed to add to waterpH of 98.079 M H

_{2}SO_{4 }Test water pH - assume pH 7 and to reduce the pH to 6_{ }pH = - log[H

^{+}] Need pH = 6 = -log[H^{+}] and 10^{6 = }1/[H^{+}] therefore, [H^{+}] = 10^{-6}M needed Let x = L of acid added to water

(x*98.079 mol/L acid)/(x L acid + 0.07393 L water) = 10

^{-6}mol/Ltake out units,

98.079*x/(x + 0.07393) = 10

^{-6}(solve for x)98.079*x = 10

^{-6}*(x + 0.07393) = 10^{-6}*x + 0.007393 (combine like terms)98.079*x = 0.007393

x = 0.007393/98.079 = 7.5378 x 10

^{-5}L = 7.5378 x 10^{-2}mL acidor 1 eye drop/(0.05 mL) x 7.5378 x 10

^{-2}mL acid = 1.5 eye drops 98.079 M acid needed to add to 2.5 fl oz waterTo raise 2.5 oz of water (at pH 7) to pH 8