Homework Answers
Answer
1.0
Explanation
H3O+(aq) + OH-(aq) -------> H2O(l)
Stoichiometrically, 1mole of OH- reacts with 1mole of H3O+
moles of OH- consumed = (1.0mol/1000ml) ×5 ml = 0.005mol
moles of H3O+ present in the original solution = 0.005mol
Concentration of H3O+ in the original solution = (0.005mol/50ml)×1000ml = 0.1M
pH = -log[H3O+]
pH = -log(0.1M)
pH = 1.0
Add Answer to:
50.0 mL of H30+ is diluted with 50.0 mL of water and titrated to the equivalence...
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point (i.e. 50.0 mL EDTA has been added), what is the equilibrium concentration of Ca2, [Ca2...
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point (i.e. 50.0 mL EDTA has been added), what is the equilibrium concentration of Ca2, [Ca2], and what is the pCa? The formation constant of CaY, Kaa 5.0 x 1010, and α4 of EDTA at pH 10.0 is 0.35
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point...
50.0 mL of 0.150 M HNO2 is titrated to its equivalence point with 1.00 M NaOH....
50.0 mL of 0.150 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
50.0 mL of 0.275 M HNO2 is titrated to its equivalence point with 1.00 M NaOH....
50.0 mL of 0.275 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point? Ka for HNO2 is 4.0x10^-4
A 50.0 mL sample of 0.25 M formic acid (HCOOH) aqueous solution is titrated with 0.125...
A 50.0 mL sample of 0.25 M formic acid (HCOOH) aqueous solution is titrated with 0.125 M NaOH solution. Ka of HCOOH = 1.7 x 10−4. a. Calculate the pH of the solution after 50 mL of NaOH solution has been added. b. How many mL of 0.125 M NaOH need to be added to the sample to reach the equivalence point? What is the pH at the equivalence point?
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the...
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.773 Volume of NaOH required to reach endpoint: (ml) 19.0 pH of the mixture Ihalf neutralized solution 3.54 Calculate the following...
4.42 mL of water sample was diluted to 101.2 mL with DI water and titrated with...
4.42 mL of water sample was diluted to 101.2 mL with DI water and titrated with a 0.01 M EDTA solution. The end point determined after addition 21.73 mL of 0.01 M EDTA. Determine the "hardness' of the water sample (represented at "Ca^2+" in ppm). A blank titration required 0.35 mL of EDTA.
A 0.898 grain sample of an unknown monoprotic acid is dissolved in 50.0 mL of water...
A 0.898 grain sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a 0.372 M aqueous sodium hydroxide solution. It is observed that after 15.0 milliliters of sodium hydroxide have been added, the pH is 5.053 and that an additional 6.90 mL of the sodium hydroxide solution is required to reach the equivalence point. What is the molecular weight of the acid? g/mol What is the value of K_a for the acid?
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10. A 50.0 mL sample of water containing both Ca? and Mg is titrated in another 50.0 mL sample, the Mg* was precipitated as Mg(OH)2...
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10. A 50.0 mL sample of water containing both Ca? and Mg is titrated in another 50.0 mL sample, the Mg* was precipitated as Mg(OH)2 and then, Ca2* was titrated at pH 13 with 9.25 ml of the same EDTA solution. Calculate ppm CaCo, (FW-100.09) and MgCO, (FW-84.31) in the sample. FWca: 40.08; FWm: 24.30
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10....
A 50.0 mL solution of 0.174 M potassium alaninate (H2NC2H5CO2K) is titrated with 0.174 M HCl....
A 50.0 mL solution of 0.174 M potassium alaninate (H2NC2H5CO2K) is titrated with 0.174 M HCl. The pKa values for the amino acid alanine are 2.344 (pKa1) and 9.868 (pKa2), which correspond to the carboxylic acid and amino groups, respectively. a. Calculate the pH at the first equivalence point b. Calculate the pH at the 2nd equivalence point
4. What is the pH of 2.00mL of 1.0 M HCI solution diluted to 50.0 mL?...
4. What is the pH of 2.00mL of 1.0 M HCI solution diluted to 50.0 mL? ( 2pts)
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point (i.e. 50.0 mL EDTA has been added), what is the equilibrium concentration of Ca2, [Ca2], and what is the pCa? The formation constant of CaY, Kaa 5.0 x 1010, and α4 of EDTA at pH 10.0 is 0.35
4. 50.0 mL of 0.00200 M Ca2 solution buffer at pH 10.0 is titrated with 0.00200 M EDTA. At equivalence point...
A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured. Mass of acid weighed out (grams) 0.773 Volume of NaOH required to reach endpoint: (ml) 19.0 pH of the mixture Ihalf neutralized solution 3.54 Calculate the following...
A 0.898 grain sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a 0.372 M aqueous sodium hydroxide solution. It is observed that after 15.0 milliliters of sodium hydroxide have been added, the pH is 5.053 and that an additional 6.90 mL of the sodium hydroxide solution is required to reach the equivalence point. What is the molecular weight of the acid? g/mol What is the value of K_a for the acid?
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10. A 50.0 mL sample of water containing both Ca? and Mg is titrated in another 50.0 mL sample, the Mg* was precipitated as Mg(OH)2 and then, Ca2* was titrated at pH 13 with 9.25 ml of the same EDTA solution. Calculate ppm CaCo, (FW-100.09) and MgCO, (FW-84.31) in the sample. FWca: 40.08; FWm: 24.30
Problem 4: EDTA titration with 16.55 ml of o.0114 M EDTA at pH-10....
4. What is the pH of 2.00mL of 1.0 M HCI solution diluted to 50.0 mL? ( 2pts)
Most questions answered within 3 hours.
-
Where is the error in this code sequence?
String s1 = "Hello";
String s2 = "ello";...
asked 11 months ago -
Financial data for Joel de Paris, Inc., for last year
follow:
Joel de Paris, Inc.
Balance...
asked 11 months ago -
Consider this reaction:
Al2(SO4)3 (aq)+ BaCl3
(aq) Al2Cl6 (aq)- +
3BaSO4(s) . What is the...
asked 11 months ago -
Suppose that Savneet is considering increasing her
recent random sample from 20 car rentals to 40...
asked 11 months ago -
Trucks arrive at an unloading terminal at an average rate of 120
per hour.
Trucks arrive...
asked 11 months ago -
Why are methanol and ethanol completely soluble in water while
octanol is not very little soluble....
asked 11 months ago -
A facilities manager at a university reads in a research report
that the mean amount of...
asked 11 months ago -
When the CuSO4 is rehydrated by adding water to the anhydrous
compound, is this an endothermic...
asked 11 months ago -
A ray of sunlight is passing from diamond into crown glass; the
angle of incidence is...
asked 11 months ago -
A block of mass 0.249 kg is placed on top of a light, vertical
spring of...
asked 11 months ago -
how do the kidneys compensate in the presences of acidosis
a) trigger hyperventilate
b) reserve acid...
asked 11 months ago -
Question 501 pts
The rental rate of capital to the firm increases. Which of the
following...
asked 11 months ago