6. A solution of 0.1000 M HCl was used to titrate 25.00 mL of Na2CO3 (0.0800 M). Calculate the pH of the solution when adding following amount of HCl. For H2CO3 Kal = 4.46 × 10-7 a) 0.00 mL (before the titration) x Ka2 = 4.69 × 10-11 x b) 5.00 mL c) at the first equivalent point d) 45.00 mL
6. A solution of 0.1000 M HCl was used to titrate 25.00 mL of Na2CO3 (0.0800 M). Calculate the pH of the solution when adding following amount of HCl. For H2CO3 Kal = 4.46 × 10-7 a) 0.00 mL (before the titration) x Ka2 = 4.69 × 10-11 x b) 5.00 mL c) at the first equivalent point d) 45.00 mL
Chapter7: Neutralization Titrations And Graphical Representations
Section: Chapter Questions
Problem 7P
Related questions
Question
![6.
A solution of 0.1000 M HCl was used to titrate 25.00 mL of Na2CO3 (0.0800 M).
Calculate the pH of the solution when adding following amount of HCl.
For H2CO3
Kal = 4.46 × 10-7
a) 0.00 mL (before the titration)
x
Ka2 = 4.69 × 10-11
x
b) 5.00 mL
c) at the first equivalent point
d) 45.00 mL](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb6a2865-a18b-481f-9763-08158873ce71%2F2031bb03-5b7a-47a4-99f4-986675e78728%2Fjmn59d4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:6.
A solution of 0.1000 M HCl was used to titrate 25.00 mL of Na2CO3 (0.0800 M).
Calculate the pH of the solution when adding following amount of HCl.
For H2CO3
Kal = 4.46 × 10-7
a) 0.00 mL (before the titration)
x
Ka2 = 4.69 × 10-11
x
b) 5.00 mL
c) at the first equivalent point
d) 45.00 mL
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