How many moles of NaOH must you add a 0.200 mol of benzonic acid (K_a = 6.28 tim

Question

How many moles of NaOH must you add a 0.200 mol of benzonic acid (K_a = 6.28 times 10^-5 to make 1.00 L of a pH 4.00 buffer)? a. 1.000 b. 0.025 c. 0.250 d. 0.077 A solution of Compound X has a transmittal of 15.8%.at 475 nm in a 2.00 cm cell. The molar absorptivity of Compound X at this wavelength is 5.47 times 10^4 L/mol-cm. What is the concentration of compound X in the solution? a. 1.95 times 10^-6 M b. 3.92 times l0^-6 M c. 7.32 times 10^-6 M d. 7.84 times 10^-6 M At 450 and 575 nm, the indicator species Hln and ln- have the tabulated molar absorptivities. An unknown solution of this indicator had measured absorbance of 0.120 at 450 nm and 0.040 at 575 nm in a 1.00-cm cell. Calculate the total concentration of the indicator. a. 1.03 times 10^-5 M b. 1.03 times 10^-6 M c. 3.30 times 10^-5 M d. 2.06 times l0^-5 M

Explanation / Answer

3)

let moles of NaOH added be x

then x mol of NaOH and benzoic acid will react to form x mol of sodium benzoate
remaining mol of benzoic acid = (0.200-x) mol
since volume is 1 L, number of moles will be same as concentration

pka = -log Ka
= -log (6.28*10^-5)
= 4.202

usE:
pH = pKa + log {[sodium benzoate] /[benzoic acid]}
4.00 = 4.202 + log (x / (0.200-x))
log (x / (0.200-x)) = -0.202
x/(0.200-x) = 0.628
x = 0.1256 - 0.628*x
x = 0.077 mol

Answer: d

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