A titration experiment is done with 1.37E-2 M NaOH to determine the acid concentration in 0.631 L...

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A titration experiment is done with 1.37E-2 MNaOH to determine the acid concentration in 0.631L of an unknown acidic solution. It was found that to reach theequivalence point 12.3 mL of strong base wasneeded. What was the pH of the original acidic solution?

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By Molarity equation, M1V1 ( NaOH) = M2V2( unknown acid)Â Â 1.37*10^-2 * 12.3ml = M2* 631ml.Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â So, M2= 2.6*10^-3.Â Â Â Â Â Â Concentration of acid= 2.6*10^-3. So, pH= -log[H3O+]= - log[ 2.6*10^-3]= 2.585.Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â l

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