Aluminum sulfate, known as cake alum, has a remarkably wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate. (a) Write balanced total and net ionic equations for its reaction with aqueous NaOH. (Type your answer using the format (NH4)2CO3 for (NH4)2CO3, [NH4]+ for NH4+, and [Ni(CN)4]2- for Ni(CN the lowest possible coefficients. Type the cation before the anion.) overall equation 1 V NAOH(aq) → 2 v Al(SO4)3 |× (aq)+ 6 x Al x (aq) + 3 x SO4 x (s) net ionic V AI x (aq) + 3 V OH Ix (aq) → 1 V AI(OH)3 v (s) 1 (b) What mass of precipitate forms when 104.5 ml of 0.616 M NAOH is added to 517 ml of a solution that contains 18.4 g aluminum sulfate per liter? 4 1.67

Aluminum sulfate, known as cake alum, has a remarkably wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate. (a) Write balanced total and net ionic equations for its reaction with aqueous NaOH. (Type your answer using the format (NH4)2CO3 for (NH4)2CO3, [NH4]+ for NH4+, and [Ni(CN)4]2- for Ni(CN the lowest possible coefficients. Type the cation before the anion.) overall equation 1 V NAOH(aq) → 2 v Al(SO4)3 |× (aq)+ 6 x Al x (aq) + 3 x SO4 x (s) net ionic V AI x (aq) + 3 V OH Ix (aq) → 1 V AI(OH)3 v (s) 1 (b) What mass of precipitate forms when 104.5 ml of 0.616 M NAOH is added to 517 ml of a solution that contains 18.4 g aluminum sulfate per liter? 4 1.67

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

(a) The sodium ion did not take part in this chemical reaction. What do we call such an ion? (b) Draw a simple diagram which shows how the sodium ion mixes with water in solution. What do we call this physical process?
What volume of 2.5 M H3PO4 (aq) would be required to neutralize 2.50 mL of 1.5 M NaOH? НЗРО4 (аq) + 3 NaOH (ag) --> Na3PО4 (aq) + 3 Н20 ()
You are preparing standard acid and base solutions for the laboratory, using potassium hydrogen phthalate (KHC₈H₄O₄, abbreviated KHP) as the primary standard. KHP (molar mass = 204.22 g/mol) has one acidic hydrogen. You prepared solutions of both NaOH and HCl. It took 22.65 mL of the NaOH solution to titrate (react exactly with) 1.55 g KHP. It then took 32.35 mL of HCl solution to titrate 25.00 mL of the NaOH solution. What is the molarity of the HCl solution?
(a) A 50.0 mL solution is prepared to be 1.29 M acetylsalicylic acid. In the first step, 8.55 mL of NaOH is titrated into the solution until the pH is exactly 5.0. What is the concentration of the titrant (NaOH)? (b) In the second step, enough 5.85 M nitric acid is added to the solution after the titration in part (a) is complete until the pH is one unit lower than the pKa of acetylsalicylic acid. What volume (mL) of nitric acid was added?
In sediments and waterlogged soil, dissolved O2 concentrations are so low that the microorganisms living there must rely on other sources of oxygen for respiration. Some bacteria can extract the oxygen in sulfate ions, reducing the sulfur in them to hydrogen sulfide gas and giving the sediments or soil a distinctive rotten-egg odor.Write the net ionic equation for the reaction under acidic conditions (H3O+) that releases O2 from sulfate and forms hydrogen sulfide gas. Use water as the reactant in the half-reaction that describes the formation of oxygen.
A titration involves adding a reactant of known quantity to a solution of an another reactant while monitoring the equilibrium concentrations. This allows one to determine the concentration of the second reactant. The equation for the reaction of a generic weak acid HA with a strong base is HA(aq) + OH(aq) →A¯(aq) + H₂O (1) A certain weak acid, HA, with a K, value of 5.61 x 10 ", is titrated with NaOH. 6 Part A A solution is made by titrating 9.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places. ▸ View Available Hint(s) pH = Submit ▾ Part B pH = IVE] ΑΣΦ Submit A → c More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 66.0 mL ? Express the pH numerically to two decimal places. ▸ View Available Hint(s) IVE ΑΣΦ A + Ċ ? Doowoon Review | Constants | Periodic Table ?
The standardized NaOH was used to determine the unknown concentration of H3PO4. The standard NaOH concentrtion is 0.126M. The standard sulfuric acid concentration is 0.150M. 3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l) Three flasks containing 25.00 mL H3PO4 each was titrated with NaOH. The data is below: Titration 1 Titration 2. Titration 3 Starting volume buret 2.62 mL 18.60 mL 1.20mL Buret volume at end 18.46 mL 34.40mL 16.84mL Calculate the concentration of H3PO4.
Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid HCl, which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HCl through this reaction: HCl(aq)+NaHCO3(aq)→NaCl(aq)+H2O(l)+CO2(g) The CO2 gas produced is what makes you burp after drinking the solution. Suppose the fluid in the stomach of a woman suffering from indigestion can be considered to be 200.mL of a 0.089M HCl solution. What mass of NaHCO3 would she need to ingest to neutralize this much HCl? Be sure your answer has the correct number of significant digits.
A solution of sodium hydroxide is standardized against potassium hydrogen phthalate (molecular mass = 204.224 g mol-1). From the following data, calculate the molarity of the NaOH solution. The mass of the KHP is 0.7840 g. Before the titration the buret reading was 0.8400 mL and afterwards it was 39.20 mL. A 85.12 mL sample of a solution of sulfuric acid, H2SO4, is neutralized by 59.82 mL of the NaOH solution from the problem above. Calculate the molarity of the sulfuric acid solution.
A pure organic compound has the formula C4H8SOx. A sample of a compound wasdecomposed; the sulfur was converted to sulfate and titrated in a modified version of theadsorption indicator procedure. From the following data, determine the correct formula forthe compound.Given: weight of sample = 12.64 mgBa(ClO4)2 needed for titration = 10.60 mL of 0.0100 M Ba(ClO4)2
A 0.5504-g sample of KHP (potassium hydrogen phthalate, KHC8H4O4; molar mass = 204.22 g/mol) is completely dissolved in enough deionized water to make 50.00 mL of solution. The titration of this solution with a KOH solution requires 20.50 mL of the base (KOH solution) to reach the “end-point”. What is the molar concentration of KOH solution? The reaction occurs as follows: KOH(aq) + KHC8H4O4(aq) --> K2C8H4O4(aq) + H2O(l) (A) 0.2457 M (B) 0.1315 M (C) 0.05390 M (D) 0.03823 M
1. a) Write the formula for the conjugate base of HPO4^-2 b) Write the formula for the conjugate acid of NH3