Question

When 220. mL of 1.05 × 10^-4 M sulfuric acid is added to 135 mL of 2.75 × 10^-4 M NaOH, the resulting solution will be

basic

acidic.

neutral.

It is impossible to tell from the information given.

Answer 1

It is impossible to say from the given information When 220. mL of 1.05 x 10-4 M sulfuric acid is added to 135 mL of 2.75 x 10-4 M NaOH, the resulting solution will be Explanation: Calculating moles of H2SO4 concentration of H2SO4 = 1.05 x 10-M Volume of H2SO4 = 220.0 mL Therefore moles of H2SO4=(220 mL) X (1.05 x 104M)=0.0231 mmol Now, calculating minol of NaOH: Given concentration of NaOH = 2.75 x 104 M Volume of NaOH added = 135 mL Therefore mmol of NaOH =(135 mL) X (2.75 x 104 M)=0.0371 mmol The new concentration of the H2SO4 and NaOH is therefore [H2SO4]final=0.0231 mmol/(220 + 135) mL = 6.5 X 10-M [NaOH]final=0.0371 mmol/(220 + 135) mL = 1.04 x 10-4M Applying the Henderson - Hasselbalch equation to find the pH. That is, pH =pKa + log([base] /[acid]) where [base]= [NaOH]final = 1.04 x 104M (acid]= [H2SO4]final = 6.5 X 10M And pKa = -log Ka As Ka value of H2SO4 is not given in question so it is impossible to know the pH with given information so we cannot say weather the solution will be acidic, basic or neutral

It is impossible to say from the given information When 220. mL of 1.05 x 10-4 M sulfuric acid is added to 135 mL of 2.75 x 10-4 M NaOH, the resulting solution will be Explanation: Calculating moles of H2SO4 concentration of H2SO4 = 1.05 x 10-M Volume of H2SO4 = 220.0 mL Therefore moles of H2SO4=(220 mL) X (1.05 x 104M)=0.0231 mmol Now, calculating minol of NaOH: Given concentration of NaOH = 2.75 x 104 M Volume of NaOH added = 135 mL Therefore mmol of NaOH =(135 mL) X (2.75 x 104 M)=0.0371 mmol The new concentration of the H2SO4 and NaOH is therefore [H2SO4]final=0.0231 mmol/(220 + 135) mL = 6.5 X 10-M [NaOH]final=0.0371 mmol/(220 + 135) mL = 1.04 x 10-4M Applying the Henderson - Hasselbalch equation to find the pH. That is, pH =pKa + log([base] /[acid]) where [base]= [NaOH]final = 1.04 x 104M (acid]= [H2SO4]final = 6.5 X 10M And pKa = -log Ka As Ka value of H2SO4 is not given in question so it is impossible to know the pH with given information so we cannot say weather the solution will be acidic, basic or neutral