thank u Draw a titration curve for the titration of 35.00 mL of a 0.145 M solution of hydrobromicacid, HBr, with 0.145 M sodium hydroxide, NaOH.1,The volume of titrant required to reach the equivalence point.2,The initial pH of the analyte solution3, The pH of the analyte solution after the addition of 5.00 mL of titrant4,The pH of the analyte solution after the addition of 15.00 mL of titrant5, The pH of the analyte solution at the equivalence point6, The pH of the analyte solution after the addition of 35.00 mL of titrant

Answered: Draw a titration curve for the…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Draw a titration curve for the titration of 35.00 mL of a 0.145 M solution of hydrobromic
acid, HBr, with 0.145 M sodium hydroxide, NaOH.
1,The volume of titrant required to reach the equivalence point.
2,The initial pH of the analyte solution
3, The pH of the analyte solution after the addition of 5.00 mL of titrant
4,The pH of the analyte solution after the addition of 15.00 mL of titrant
5, The pH of the analyte solution at the equivalence point
6, The pH of the analyte solution after the addition of 35.00 mL of titrant

Expert Solution

Step 1

$T h e p H o f a s o l u t i o n i s g u i d e d b y t h e c o n c e n t r a t i o n o f t h e h y d r o g e n i o n s . p H = - \log_{10} [H^{+}] w h e n a c i d a n d b a s e a r e m i x e d, t h e a c t u a l r e a c t i o n o c c u r s b e t w e e n t h e h y d r o g e n i o n a n d h y d r o x i d e i o n . A t e q u i v a l e n c e p o i n t, t h e m o l e s o f a c i d a n d b a s e r e a c t e d a r e e q u a l .$

Step 2

$\begin{array}{rcl} 1 . h e n c e, a t e q u i v a l e n c e p o i n t, \\ M_{a} V_{a} & = & M_{b} V_{b} \\ 35 m L \times 0.145 M & = & 0.145 M \times V_{b} \\ V_{b} i s 35 m L \\ I t m e a n s 35 m L o f N a O H a r e n e e d e d f o r e q u i v a l e n c e p o i n t . \\ 2 . i n i t i a l l y o n l y a c i d i s p r e s e n t, s o [H^{+}] & = & 0.145 M \\ p H & = & - \log_{10} (0.145) \\ 0.84 \\ t h e i n i t i a l p H i s 0.84 \end{array}$

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