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

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?

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Expert 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

queen_honey_blossom answered 1 year ago
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