A solution of tartaric acid (H:C4H.Oe) With a knówn concentration of 0.155 M H:C.H.O. is titrated with a 0.425 M NaOH solution. How many mL of NaOH are required to reach the second equivalence point with a starting volume of 70.0 mL H:CaH.O., according to the following balanced chemical equation: H:C.H.Os + 2 NaOH – Na:C.H.Os + 2 H:0

Chemistry
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Chapter15: Acid-base Equilibria
Section: Chapter Questions
Problem 132IP: A 10.00-g sample of the ionic compound NaA, where A is the anion of a weak acid, was dissolved in...
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A solution of tartaric acid (H:C&H.Os) with a known concentration of 0.155
M H2C«H«Os is titrated with a 0.425 M NaOH solution. How many mL of NaOH are
required to reach the second equivalence point with a starting volume of 70.0
mL H.C«H«Os , according to the following balanced chemical equation:
H:CH.Os +
2
NAOH - Na:CaH.Os +
2
H:O
STARTING AMOUNT
ADD FACTOR
ANSWER
RESET
*( )
1000
0.425
0.001
0.0217
12.8
25.5
0.155
1
0.0511
51.1
5.11 x 104
70.0
g NaOH
L H2CsH.Os
mol H:CAH.O6
M H2CH«Os
mol NaOH
mL NaOH
M NaOH
mL H2CaH«Os
L NAOH
g H2C4H.O6
1:35 PM
P Type here to search
46°F
2/27/2022
Transcribed Image Text:101 Chem101 b My Questions | bartleby А аpp.101edu.co E Apps e Resource Library -. Dashboard 101 Chem101 E Reading list Word wco Writing Centers at. 11-0218 NCI Theor.. Question 5 of 11 Submit A solution of tartaric acid (H:C&H.Os) with a known concentration of 0.155 M H2C«H«Os is titrated with a 0.425 M NaOH solution. How many mL of NaOH are required to reach the second equivalence point with a starting volume of 70.0 mL H.C«H«Os , according to the following balanced chemical equation: H:CH.Os + 2 NAOH - Na:CaH.Os + 2 H:O STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 1000 0.425 0.001 0.0217 12.8 25.5 0.155 1 0.0511 51.1 5.11 x 104 70.0 g NaOH L H2CsH.Os mol H:CAH.O6 M H2CH«Os mol NaOH mL NaOH M NaOH mL H2CaH«Os L NAOH g H2C4H.O6 1:35 PM P Type here to search 46°F 2/27/2022
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