Question

К 200 мл 0,5 М раствора ортофосфата натрия добавили некоторый объем 1 М соляной кислоты и получили раствор, содержащий три соли, причём все три - в равных количествах (моль). Сколько миллилитров кислоты добавили? Какой соли из приведённых в таблице НЕ БЫЛО в полученном растворе? (гидролиз не учитывать). Объём раствора HCl(мл)- ? Какая соль отсутствовала? Na3PO4, Na2HPO4, NaH2PO4, NaCl

Answer 1

Ответ:

Знал бы я сам...

Объяснение:

Answer 2

Ответ:

Объём раствора HCl (мл) – 1

Какая соль отсутствовала?  Na2HPO4

Объяснение:

Answer 3

Question Analysis

The question is asking for the following information: 1. The volume of hydrochloric acid added to a solution containing 200 ml of 0.5 M sodium orthophosphate and resulting in a solution containing three salts in equal amounts. 2. The salt that was not present in the resulting solution.

Let's address each part of the question separately.

Volume of Hydrochloric Acid Added

To determine the volume of hydrochloric acid added, we need to consider the reaction between sodium orthophosphate (Na3PO4) and hydrochloric acid (HCl). The balanced chemical equation for the reaction is:

3Na3PO4 + 6HCl → 3NaCl + H3PO4

From the equation, we can see that for every 6 moles of hydrochloric acid, 1 mole of phosphoric acid (H3PO4) is produced. Since the question states that all three salts are present in equal amounts, it means that 1 mole of each salt is present in the resulting solution. Therefore, we can conclude that 1 mole of phosphoric acid is present in the solution.

To calculate the volume of hydrochloric acid added, we need to determine the number of moles of hydrochloric acid required to produce 1 mole of phosphoric acid. Since the molar ratio between hydrochloric acid and phosphoric acid is 6:1, we can conclude that 6 moles of hydrochloric acid are required to produce 1 mole of phosphoric acid.

Given that the initial solution contains 200 ml of 0.5 M sodium orthophosphate, we can calculate the number of moles of sodium orthophosphate:

Number of moles = Volume (in liters) × Concentration (in moles/liter) Number of moles = 0.2 L × 0.5 mol/L Number of moles = 0.1 mol

Since the number of moles of sodium orthophosphate is equal to the number of moles of phosphoric acid, we can conclude that 0.1 mol of phosphoric acid is present in the solution.

To calculate the volume of hydrochloric acid added, we can use the following equation:

Volume of hydrochloric acid (in liters) = Number of moles of hydrochloric acid × Molarity of hydrochloric acid

Since we know that 6 moles of hydrochloric acid are required to produce 1 mole of phosphoric acid, we can calculate the volume of hydrochloric acid added:

Volume of hydrochloric acid (in liters) = (0.1 mol phosphoric acid) × (6 mol hydrochloric acid / 1 mol phosphoric acid) × (1 L / 1,000 ml)

Let's calculate the volume of hydrochloric acid added:

Volume of hydrochloric acid (in liters) = 0.1 × 6 × 0.001 = 0.0006 L

To convert the volume to milliliters, we multiply by 1,000:

Volume of hydrochloric acid (in milliliters) = 0.0006 L × 1,000 = 0.6 ml

Therefore, 0.6 milliliters of hydrochloric acid were added to the solution.

Salt Not Present in the Resulting Solution

The question asks which of the salts listed in the table was not present in the resulting solution. The salts listed are Na3PO4, Na2HPO4, NaH2PO4, and NaCl.

Since the resulting solution contains three salts in equal amounts, and we have already determined that 1 mole of each salt is present, we can conclude that all four salts were present in the resulting solution.

Therefore, none of the salts listed in the table was absent in the resulting solution.

Volume of HCl in the Solution

The question does not provide enough information to determine the volume of hydrochloric acid in the resulting solution. We only know that 0.6 milliliters of hydrochloric acid were added, but we don't know the final volume of the solution or the concentration of the hydrochloric acid.

Therefore, we cannot determine the volume of hydrochloric acid (HCl) in the solution based on the information provided.

To summarize: - The volume of hydrochloric acid added was 0.6 milliliters. - All four salts listed in the table (Na3PO4, Na2HPO4, NaH2PO4, and NaCl) were present in the resulting solution. - The volume of hydrochloric acid (HCl) in the solution cannot be determined based on the information provided.