Your shopping cart is empty!
Explore Related Concepts


acid base titration calculator
Best Results From Yahoo Answers Youtube
From Yahoo Answers
Question:Check my answers please!
1. A 0.484g sample of impure NH4Br is treated with 25.00 mL of 0.2050M NaOH and heated to drive off the NH3. The unreacted NaOH in the reaction mixture after heating required 9.95 mL of 0.0525M H2C2O4 to neutralize.
a. How many grams of NH4Br were in the original sample?
H2C2O4(aq) + 2NaOH(aq) > Na2C2O4 + 2H2O(l)
5.13x10^3 mol NaOH added
5.22x10^4mol H2C2O4 x 2 mol NaOH/1 mol H2C2O4 = 1.04x10^3 mol NaOH unreacted
5.13x10^3  1.04x10^3 = 4.09x10^3 mol NH4Br x 97.9g NH4Br/1mol NH4Br
= .401g NH4Br
b. What is the percentage NH4Br in the original sample?
.401g/.484g x 100%
= 82.9% NH4Br in the sample
Answers:Your method is correct, but the exact values I think are not. When doing calculations, save rounding till the end and instead track your sig figs with a line underneath the correct digit. Without rounding during the calculations, you would use 5.125x10^3 and 1.04475x10^3. This adds up to a different answer at the end. For the answers I got .400g and 82.6%.
Answers:Your method is correct, but the exact values I think are not. When doing calculations, save rounding till the end and instead track your sig figs with a line underneath the correct digit. Without rounding during the calculations, you would use 5.125x10^3 and 1.04475x10^3. This adds up to a different answer at the end. For the answers I got .400g and 82.6%.
Question:A 500 mL aqueous solution contains 1.36 g of HCl. What is the pH?
Give your answer to 3 significant figures.
Answers:MW of HCl: 36.463 (1.36 g)(1 mole / 36.463 g) = 0.037298 moles Concentration = (0.037298 moles / 0.500 L) Concentration = 7.4596 x 10^2 M And the reaction is: HCl > H+ + Cl And since it is assumed that HCl completely disassociates And also since H+ is produced in a 1:1 ratio of HCl that disassociates [H+] = 7.4596 x 10^2 M pH =  log [H+] pH = 1.13
Answers:MW of HCl: 36.463 (1.36 g)(1 mole / 36.463 g) = 0.037298 moles Concentration = (0.037298 moles / 0.500 L) Concentration = 7.4596 x 10^2 M And the reaction is: HCl > H+ + Cl And since it is assumed that HCl completely disassociates And also since H+ is produced in a 1:1 ratio of HCl that disassociates [H+] = 7.4596 x 10^2 M pH =  log [H+] pH = 1.13
Question:i have 18.33 mL of unknown molarity HCL
I have 25.00 mL of .2 M NaOH
What do I do now?
Answers:write the balanced eq HCl + NaOH > NaCl + H2O So HCl : NaOH = 1:1 find moles OH used; n = conc. x volume = 0.2M x 0.025L = 0.005 mol ratio is 1:1 so 0.005 mol HCl neutralised conc. (molarity) = n / volume = 0.005mol / .01833L = 0.27M
Answers:write the balanced eq HCl + NaOH > NaCl + H2O So HCl : NaOH = 1:1 find moles OH used; n = conc. x volume = 0.2M x 0.025L = 0.005 mol ratio is 1:1 so 0.005 mol HCl neutralised conc. (molarity) = n / volume = 0.005mol / .01833L = 0.27M
Question:In a titration, a standard solution of hydrochloric acid (concentration 0.2 mol dm3) was added to 25 cm3 of a sodium hydroxide solution, concentration unknown. On addition of 18.5 cm3 HCl, the equivalence point was reached. What is the concentration of the sodium hydroxide solution?
a. 0.2
b. 3.7 103.
c. 0.296
d. 0.148
Answers:moles HCl = 0.2 x 0.0185 dm^3 =0.0037 moles NaOH = 0.0037 Molarity of NaOH = 0.0037/ 0.025 dm^3=0.148 M
Answers:moles HCl = 0.2 x 0.0185 dm^3 =0.0037 moles NaOH = 0.0037 Molarity of NaOH = 0.0037/ 0.025 dm^3=0.148 M
From Youtube
Acids and bases. Calculating pH. Titrations (1) :Chemistry: Acids and bases. Calculating pH and pOH. Water autoionization; water ionproduct constant (Kw). Acid dissociation constant (Ka). Buffer solutions; HendersonHasselbach equation. Titrations; titration curve; equivalence point; halfequivalence point This is a recording of a tutoring...
Acids and bases. Calculating pH. Titrations (2) :Chemistry: Acids and bases. Calculating pH and pOH. Water autoionization; water ionproduct constant (Kw). Acid dissociation constant (Ka). Buffer solutions; HendersonHasselbach equation. Titrations; titration curve; equivalence point; halfequivalence point This is a recording of a tutoring...