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need help answering question 4, it requires you to use question 3 that is boxed in pink. thank you.
3. A student following the procedure in this experiment prepared three solutions by adding 11.50,23.00, and 34.50 mL of a 7.60x 10 M NaOH solution to 50.00 mL of a 0.112M solution of a weak acid. The solutions were labeled 1, 2, and 3, respectively. Each of the solutions was diluted to a total volume of 100.0 mL with distilled water. The pH readings of these solutions were: (1) 6.42, (2) 6.81, and (3) 7.09 (1) Convert the pH of each solution to an equivalent HO ion concentration 0 2) 10 (2) For each solution, calculate the number of moles of acid added. (3) For each solution, calculate the number of moles of OH ion a dded 8.00087 nde 3( 23.00a/1000)(7.60×10-211) 0.00伃48t mole (4) For each solution, determine the number of moles of HAn and of An ion at equilibrium. En solcha! 2 . Moles of HAn = mcts of An, = rnoles of base : 1.75-10°, cles 40 xl”. (.75 x1D) = 3.85 x 10-3 roles ‘ Volurie : 100 rt –
Expert Answer
4) The student forgot to open the stopcock, so the burette tip remained empty. When he titrate NaOH with acid, it will appear that more NaOH had been delivered to the acid sample than what actually was because the air trapped in the burette was accounted for as NaOH solution. 7.60 x 10-2 M NaOH solution neutralizes less volume of acid. So, concentration of acid will be high, which can be calculated by using formula Nb x Vb = NaVa. More the concentration of acid, greater the value of Ka will be.
The value of Ka will be more, which is in question 3
