In: Chemistry
A titration experiment is done with 1.37E-2 M NaOH to determine the acid concentration in 0.631 L of an unknown acidic solution. It was found that to reach the equivalence point 12.3 mL of strong base was needed. What was the pH of the original acidic solution?
By Molarity equation, M1V1 ( NaOH) = M2V2( unknown acid) 1.37*10^-2 * 12.3ml = M2* 631ml. So, M2= 2.6*10^-3. Concentration of acid= 2.6*10^-3. So, pH= -log[H3O+]= - log[ 2.6*10^-3]= 2.585. l