HNO3 + NaOH ==> NaNO3 + HOH
Notice ratio acid/base = 1:1
pH 3.78 = (H^+) = (HNO3) = 1.7E-4 (approximate--you should do it more accurately.)
1.7E-4 moles/L x 1.6E4L = about 3 moles = moles HNO3 = moles NaOH needed.
M NaOH = moles NaOH/L NaOH. You have M and moles, solve for L.
A storage tank containing 1.6x10^4 Liters of water is found to be contaminated with nitric acid and therefore has a pH of 3.78. Calculate the volume of 0.050 M sodium hydroxide required to be added to the tank in order to neutralize (pH = 7.0) the contaminated water.
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