How many grams of NaC3H6O3 should be dissolved in 500.0 ml of 0.100 M HC3H6O3 (lactic acid) to make a buffer with a pH of 2.90? Assume there is no change in volume The image provides dissociation constants for various acids and bases, as well as solubility products for specific compounds. These values are critical in understanding chemical equilibrium in aqueous solutions.**Acid Dissociation Constants (\(K_a\)):**- HF: \(K_a = 7.1 \times 10^{-4}\)- HNO₂: \(K_a = 4.5 \times 10^{-4}\)- HCOOH (Formic acid): \(K_a = 1.7 \times 10^{-4}\)- HC₃H₆O₃: \(K_a = 1.4 \times 10^{-4}\)- CH₃CH₂COOH: \(K_a = 1.3 \times 10^{-5}\)- CH₃COOH (Acetic acid): \(K_a = 1.8 \times 10^{-5}\)- C₆H₅COOH (Benzoic acid): \(K_a = 6.3 \times 10^{-5}\)- HCN: \(K_a = 4.9 \times 10^{-10}\)**Base Dissociation Constants (\(K_b\)):**- CH₃NH₂: \(K_b = 4.4 \times 10^{-4}\)- NH₃: \(K_b = 1.8 \times 10^{-5}\)- C₅H₅N: \(K_b = 1.7 \times 10^{-9}\)**Solubility Products (\(K_{sp}\)):**- PbCl₂: \(K_{sp} = 1.7 \times 10^{-5}\)- CuCl: \(K_{sp} = 1.9 \times 10^{-7}\)- Mg(OH)₂: \(K_{sp} = 1.8 \times 10^{-11}\)These constants are essential for predicting the extent of dissociation and solubility of substances in water, which is pivotal in chemical reactions and processes.

An acid buffer is a mixture of a weak acid and its salt with a strong base. It is given that pH of buffer = 2.90 Volume of lactic acid = 500 ml Concentration of lactic acid = 0.1 MFrom the given table it can be seen that Ka for lactic acid = 1.4 x 10-4 The formula to calculate pKa is given as follows: pKa=-logKa Substitute 1.4 x 10-4 for Ka to calculate pKa of lactic acid. pKa=-log1.4×10-4=3.85 Handerson and Hasselbalch's equation can be written as follows: pH=pKa+logsaltacid Here, [salt] = concentration of salt, i.e. NaC3H6O3 and [acid] = concentration of acid, i.e. HC3H6O3 The above equation can be rearranged to calculate [salt] as follows: salt=acid×10pH-pKaor NaC3H6O3=HC3H6O3×10pH-pKa Substitute 0.1 M for [HC3H6O3], 2.90 for pH, and 3.85 for pKa in the above formula to calculate the concentration of NaC3H6O3. NaC3H6O3=0.1 M×102.90-3.85=0.011 M Molarity is defined as the number of moles of solute per liter or 1000 ml solution. So 0.011 M [NaC3H6O3] means 0.011 moles of NaC3H6O3 are present in 1000 ml solution. So number of moles of NaC3H6O3 in 500 ml solution = 0.011 mol1000 ml×500 ml=0.0055 molMolar mass of NaC3H6O3 is calculated as follows: Molar mass=1×atomic mass of Na+3×atomic mass of C+6×atomic mass of H+3×atomic mass of O Substitute 23 g/mol for the atomic mass of Na, 12 g/mol for the atomic mass of C, 1 g/mol for the atomic mass of H, and 16 g/mol for the atomic mass of O in the above formula to calculate molar mass of NaC3H6O3. Molar mass of NaC3H6O3=1×23+3×12+6×1+3×16 g/mol=113 g/mol The formula for calculating number of moles of NaC3H6O3 is as follows : Number of moles of NaC3H6O3=mass of NaC3H6O3 molar mass of NaC3H6O3 This formula can be rearranged to calculate mass of NaC3H6O3 as follows: mass of NaC3H6O3=number of moles of NaC3H6O3×molar mass of NaC3H6O3 Substitute 0.0055 mol for number of moles of NaC3H6O3 and 113 g/mol for the molar mass of NaC3H6O3 in the above formula to calculate the mass of NaC3H6O3. mass of NaC3H6O3=0.0055 mol×113 g/mol=0.6215 g Hence dissolving 0.6215 g of NaC3H6O3 in 500 ml of 0.1 M lactic acid will result in a buffer of pH 2.90.