Question

Be sure to answer all parts.

A solution is made by mixing exactly 500 mL of 0.135 M NaOH with exactly 500 mL of 0.100 M CH3COOH. Calculate the equilibrium concentration of the species below. Ka of CH3COOH is 1.8 × 10^−5

a.) [H+]   Enter your answer in scientific notation.

b.) [OH−] M

c.) [CH3COOH]     Enter your answer in scientific notation.

d.) [Na+] M

e.) [CH3COO−] M

Answer 1

millimoles of NaOH = 500 x 0.135 / 1000 = 0.0675

millimoles of CH3COOH = 500 x 0.1 / 1000 = 0.05

CH3COOH + NaOH   --------------> CH3COONa + H2O

0.05                0.0675                            0                    0

0                  0.0175                          0.05

here strong base reminas : so

[OH-] - 0.0175 / (500 + 500) = 1.75 x 10^-5

[H+] = 5.71 x 10^-10 M

[OH-] = 1.75 x 10^-5 M

[CH3COOH] = 0.0 M

[Na+] = 0.0675 M

[CH3COO-] = 1.75 x 10^-5 M