You are given sodium acetate, 1m hcl, nahco3 and na2co3. determine which of these four you would need and then show calculations to make buffer ph

solution:

the given compounds are sodium acetate, 1M HCL,NaHCO₃ and Na2CO₃

pH of the buffer solution is 4.7

the value of pKa of sodium bicarbonate is 6.37

the value of pKa of acetic acid is 4.7

calculate concentration of acetic acid by using the following forumula

pH=pKa+lag[salt]/[acid]

substitute the pH and Pka values in the formula.

4.7=4.7+log[salt]/[acid]

log[salt]/[acid]=0

thus, the concentration ratio of the salt and acid should be equal to each other.

Thus, concentration of sodium acetate is 0.05M

Concentration of sodium acetate= concentration of acid

= 0.05M

Volume of the buffer solution is 100mL

The buffer solution can be prepared as 0.05M of 50mL sodium acetate will react with 0.05M of 50mL of 0.05M of HCL.

The chemical equation for neutralization of the weak base with strong can be represented as show as

CH₃COONa+HCL-->CH₃COOH+NaCL

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.