AutoSaveCHEM 212 Exam 2_ Fall 2020 - WordO Search困ffsteve MSMFileHomeInsertDesignLayoutReferencesMailingsReviewViewHelpA ShareO Comments6) 25.00 mL of 0.350 M HCOOH (Formic Acid) is titrated with 0.500 M NaOH.a) Calculate the volume of NaOH required to reach the equivalence point.b) Calculate the pH of the 25.00 mL of 0.350 M HCOOH before any base has been added.c) Calculate the pH of the solution after 10.00 mL of 0.500M NaOH has been added to 25.00mL of 0.350 M HCOOH.Page 6 of 6518 wordsD, Focus90%

Answered: 6) 25.00 mL of 0.350 M HCOOH (Formic…

Chemistry

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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

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

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CHEM 212 Exam 2_ Fall 2020 - Word
O Search
困
ff
steve M
SM
File
Home
Insert
Design
Layout
References
Mailings
Review
View
Help
A Share
O Comments
6) 25.00 mL of 0.350 M HCOOH (Formic Acid) is titrated with 0.500 M NaOH.
a) Calculate the volume of NaOH required to reach the equivalence point.
b) Calculate the pH of the 25.00 mL of 0.350 M HCOOH before any base has been added.
c) Calculate the pH of the solution after 10.00 mL of 0.500M NaOH has been added to 25.00
mL of 0.350 M HCOOH.
Page 6 of 6
518 words
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90%

Expert Solution

Step 1

$\begin{array}{rcl} G i v e n d a t a, \\ M o l a r i t y o f H C O O H & = & 0.350 M \\ V o l u m e o f H C O O H & = & 12.0 m L \\ M o l a r i t y o f N a O H & = & 0.500 m L \end{array}$

Step 2

Ans a )

$\begin{array}{rcl} H C O O H + N a O H \overset{}{\to} H C O O N a + H_{2} O \\ A c c o r d i n g t o t h e r e a c t i o n, a t e q u i v a l e n c e p o i n t \\ M 1 V 1 (N a O H) & = & M 2 V 2 (H C O O H) \\ V 1 & = & \frac{M 2 V 2}{M 1} \\ V 1 & = & \frac{0.350 M \times 12.0 m L}{0.500 M} \\ = & 8.40 m L \end{array}$

Ans b)

$\begin{array}{rcl} H C O O H \overset{}{\to} H C O O^{-} + H^{+} \\ I n i t i a l 0.350 - - \\ E q u i l i b r i u m 0.350 - x x x \\ K a & = & \frac{[H C O O^{-}] [H^{+}]}{[H C O O H]} \\ 1.8 \times 10^{- 4} & = & \frac{x^{2}}{0.350 - x} \\ x c a n b e n e g l e c t e d \\ x^{2} & = & 0.350 \times 1.8 \times 10^{- 4} \\ x^{2} & = & 0.63 \times 10^{- 4} \\ x & = & \sqrt{0.63 \times 10^{- 4}} \\ = & 0.793 \times 10^{- 2} \\ p H & = & - \log [H^{+}] \\ = & - \log (0.793 \times 10^{- 2}) \\ = & 2.1 \end{array}$

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