answer

watching

views

13 Dec 2019

A 50.0 mL sample of 0.250 M sodium carbonate (Ka1 = 4.3 x10ï¿½7,Ka2 = 4.8 x 10ï¿½11) is titrated
with 0.100 M hydrochloric acid (Ka = ~102). Calculate the pH forthe following points on the titration curve.

A) 0.00 mL HCl added.

B) 50.0 mL HCl added.

C) The first equivalence point.

D) The second equivalence point.

answer

watching

views

For unlimited access to Homework Help, a Homework+ subscription is required.

You have 0 free answers left.
Get unlimited access to 3.8 million step-by-step answers.

Get unlimited access

Already have an account? Log in

Trinidad TremblayLv2

17 Dec 2019

Unlock all answers

Get 1 free homework help answer.

Already have an account? Log in

Join us on Discord

Chemistry Study Group

Join now

Related questions

Carbonic acid (H2CO3) is a weak diprotic acid with Ka1=4.43Ã10â7 and Ka2=4.73Ã10â11. When sodium bicarbonate (NaHCO3) is titrated with hydrochloric acid (HCl), it acts as a weak base according to the equation

NaHCO3(aq)+HCl(aq)âH2CO3(aq)+NaCl(aq)

Suitable indicators are those that change color within the pH range for the equivalence point of a specific titration. The expected pH at the equivalence point can be calculated using pKa values. Suitable indicators for use in titrating carbonic acid or carbonate solutions are methyl orange and phenolphthalein.

Part A

What volume of 0.130 M HCl is required for the complete neutralization of 2.00 g of NaHCO3(sodium bicarbonate)?

What volume of 0.170 M HCl is required for the complete neutralization of 1.50 g of Na2CO3(sodium carbonate)?

A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.190 g of this sample requires 23.98 mL of 0.100 MHCl. An additional 0.700 mL of 0.100 M HCl is required to reach the methyl orange endpoint. What is the percentage of Na2CO3 by mass in the sample?

a). Sketch the titration curve that is expected when a 10.00 mL mixture of 0.0100 M HNO3 and 0.0100 M diprotic oxalic acid H2C2O4 are both titrated to a final equivalence point.

The Ka's for oxalic acid are Ka1 = 0.0540; Ka2 = 5.4 x 10â5.

The titrant is 0.010 M NaOH.

b). Calculate pH at the following points in the titration and clearly indicate these points on your sketch.

(a) Starting point, 0.00 mL titrant added. (Hint: the first H+ on oxalic acid is "not-so-weak")

(b) The first equivalence point. (Hint: What species is/are present? How many mL titrant added?)

(d) The pH when a 10.0 mL excess is added, beyond the final equivalence point. (Hint: how many equivalence points should be observed?)

1.) Watch the animation, and observe the titration process using a standard 0.100 M sodium hydroxide solution to titrate 50.0 mL of a 0.100 M hydrochloric acid solution. Identify which of the following statements regarding acid-base titrations are correct

Drag items A-F to: Before equivalence point, At equivalence point, After equivalence point

A.) The pH of the solution is close to 2

B.) The pH of the solution is close to 12

C.) The pH of the solution is 7

D.) The color of the solution is pink

E.) The color of the solution is blue

F.) The pH changes most rapidly

2.) Consider the balanced equation for the titration of an acid, HCl, with a base, NaOH:

HCl(aq)+NaOH(aq)âNaCl(aq)+H2O(aq)

According to the balanced equation, to neutralize one mole of the acid (HCl) you will have to add one mole of the base (NaOH). Based on the number of moles of HCl and NaOH at a particular stage of titration, the progress of the reaction with respect to the equivalence point can be decided, as shown in the following table:

You can use the following equation to determine the number of moles in a given solution or volume of titrant: number of moles=molarity (mol/L)Ãvolume (in liters)

A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point.

Drag items A-F to Before equivalence point, At equivalence point, After equivalence point

A.) 50.0 mL of 1.00 M NaOH

B.) 150 mL of 1.00 M NaOH

C.) 200 mL of 1.00 M NaOH

D.) 10.0 mL of 1.00 M NaOH

E.) 5.00 mL of 1.00 M NaOH

F.) 100 mL of 1.00 M NaOH

Weekly leaderboard

Home

Homework Help 3,900,000

Chemistry 830,000

Unlock all answers

Related textbook solutions

Basic Chemistry

Principles of Chemistry Molecular Approach

Principles of Chemistry Molecular Approach

Chemistry: Structure and Properties

Chemistry: A Molecular Approach

Chemistry: A Molecular Approach

Principles of Chemistry: A Molecular Approach

Chemistry: The Central Science

Related questions

Weekly leaderboard