Calculate the (H+) and pH of a 2.55 x 10-4 M butanoic acid solution. The Ką of butanoic acid is 1.52 x 10-5. 0.00006324...

Question

Question

Calculate the (H+) and pH of a 2.55 x 10-4 M butanoic acid solution. The Ką of butanoic acid is 1.52 x 10-5. 0.000063246 M pH

Answer 1

Answer #1

Butanoic acid is an organic acid and hence it is a weak acid.

Weak acid is not dissociated completely . Thus the way to determine the pH and the concentration of H⁺ ions of weak acid are shown below.

The concentration of H⁺ ions = 6.225×10^—5M .

pH of the solution = 4.2

Butonsi Dissociation Cantant Ka Sinu Thus Kat TH+ Ka of butemoi auiL = 1.52 16 DButensie d M. 2.55*10 -5 (O 52 2.55M10 6.2257

Answer 2

Similar Homework Help Questions

Calculate the [H+) and pH of a 3.75 x 10-4 M butanoic acid solution. The K,...

Calculate the [H+) and pH of a 3.75 x 10-4 M butanoic acid solution. The K, of butanoic acid is 1.52 x 10-5. M pH =
Calculate the [H+] and pH of a 0.000143 M butanoic acid solution. Keep in mind that...

Calculate the [H+] and pH of a 0.000143 M butanoic acid solution. Keep in mind that the Ka of butanoic acid is 1.52 x10-5. this is a problem that will require you to use the quadratic. Start by setting up the quadratic equation before using the quadratic formula and find the coefficients. Then solve the quadratic. Quadratic formula:ax2+bx+c=0; enter the values of a, b, and c a= b= c= [H+]= M pH=
A 1.37 L buffer solution consists of 0.153 M butanoic acid and 0.326 M sodium butanoate....

A 1.37 L buffer solution consists of 0.153 M butanoic acid and 0.326 M sodium butanoate. Calculate the pH of the solution following the addition of 0.071 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52 x 10-5. pH =

A 1.41 L buffer solution consists of 0.154 M butanoic acid and 0.280 M sodium butanoate....

A 1.41 L buffer solution consists of 0.154 M butanoic acid and 0.280 M sodium butanoate. Calculate the pH of the solution following the addition of 0.068 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The K, of butanoic acid is 1.52 x 10-5. pH =
A 1.35 L buffer solution consists of 0.115 M butanoic acid and 0.331 M sodium butanoate....

A 1.35 L buffer solution consists of 0.115 M butanoic acid and 0.331 M sodium butanoate. Calculate the pH of the solution following the addition of 0.079 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The K, of butanoic acid is 1.52 x 10-5. pH =
A 1.46 L buffer solution consists of 0.108 M butanoic acid and 0.303 M sodium butanoate....

A 1.46 L buffer solution consists of 0.108 M butanoic acid and 0.303 M sodium butanoate. Calculate the pH of the solution following the addition of 0.073 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The K, of butanoic acid is 1.52 x 10-5 pH = 5.26 Incorrect

Calculate the (H+) and pH of a 2.37 x 10-4 M iodoacetic acid solution. The K,...

Calculate the (H+) and pH of a 2.37 x 10-4 M iodoacetic acid solution. The K, of iodoacetic acid is 6.68 x 10-5. [H+) = M pH =
Calculate the [H+] and pH of a 1.23 x 10-4 M nitrous acid solution. The K,...

Calculate the [H+] and pH of a 1.23 x 10-4 M nitrous acid solution. The K, of nitrous acid is 7.10 x 10-5. [H+] = 1.4 X10-4 pH = 3.9
A 1.36 L buffer solution consists of 0.157 M butanoic acid and 0.311 M sodium butanoate....

A 1.36 L buffer solution consists of 0.157 M butanoic acid and 0.311 M sodium butanoate. Calculate the pH of the solution following the addition of 0.070 moles of NaOH. Assume that any contribution of the NaOH to the volume of the solution is negligible. The ?a of butanoic acid is 1.52×10−5.

A 1.421.42 L buffer solution consists of 0.1230.123 M butanoic acid and 0.3050.305 M sodium butanoate....

A 1.421.42 L buffer solution consists of 0.1230.123 M butanoic acid and 0.3050.305 M sodium butanoate. Calculate the pH of the solution following the addition of 0.0630.063 moles of NaOHNaOH. Assume that any contribution of the NaOHNaOH to the volume of the solution is negligible. The ?aKa of butanoic acid is 1.52×10−51.52×10−5.