75mL of a 0.111 mol L-1 solution of butanoic acid(H3CCH2CH2C(=O)OH), and 38 mL of a solution...

75mL of a 0.111 mol L^-1 solution of butanoic acid(H₃CCH₂CH₂C(=O)OH), and 38 mL of a solution of KOH of molarity 0.101 mol L^-1 are mixed in a 250 mL beaker. What would be the resulting solution's pH?

Solutions

Expert Solution

queen_honey_blossom answered 9 months ago

Next > < Previous

Related Solutions

50.0 mL of 2.00 mol/L HNO3 solution and 50.0 mL of 1.00 mol/L NaOH solution, both...

50.0 mL of 2.00 mol/L HNO3 solution and 50.0 mL of 1.00 mol/L NaOH solution, both at 20.0 degree Celsius, were mixed in a calorimeter. Calculate the molar heat of neutralization of HNO3 in kJ/mol if: (1) final temperature was 28.9 degree Celsius; (2) the mass of the overall solution was 102.0 g; (3) the heat capacity of the calorimeter was 25.0 J/C; (4) and assume that the specific heat of solution is the same as water, 4.184 J/(g C);

What is the pH of a 0.16 mol/L OH- solution? 1.What is the concentration of an...

What is the pH of a 0.16 mol/L OH- solution? 1.What is the concentration of an aqueous solution of NaOH (a strong base) which has a pH of 10.1? 2.What is the concentration of an aqueous solution of Ca(OH)2 (a strong base) which has a pH of 12.05? 3.Lactic acid is a weak acid with one acidic proton. A 0.10 M solution of this acid has a pH of 2.44. What is the Ka value of this acid? 4.Hydroflouric acid,...

Consider the titration of 50.0 mL of a 0.0200 M solution of butanoic acid with 0.100...

Consider the titration of 50.0 mL of a 0.0200 M solution of butanoic acid with 0.100 M NaOH. First, calculate the equivalence point (Ve). Then, calculate the pH at the following points along the titration curve. Assume fractions are exact numbers and ignore activities for all calculations in this problem. A) VNaOH= 0.000 mL B) VNaOH= 1/2Ve C) VNaOH= 4/5Ve D) VNaOH= Ve E) VNaOH= 3/2Ve

Find the pH of a 30.0 ml butanoic acid ( HA) solution ( 0.600 M )...

Find the pH of a 30.0 ml butanoic acid ( HA) solution ( 0.600 M ) when you add the following volume of 0.600 M NaOH ( 5 points ). Ka of HA = 1.5 x 10-5 (a) 0.0 ml (b) 10.0 ml (c) 15.0 ml (d) 30.0 ml (e) 50.0 ml

A 12.9 mL solution of 0.100 mol L-1 HOCl is titrated using 0.150 mol L-1 NaOH....

A 12.9 mL solution of 0.100 mol L-1 HOCl is titrated using 0.150 mol L-1 NaOH. What is the pH of the solution after 5.18 mL of the NaOH solution is added? Express your answer to 2 decimal places. You have 5 attempts at this question. Remember you can find KA and/or KB values in your textbook in chapter 15.

A 18.6 mL solution of 0.100 mol L-1 NaOH is titrated using 0.150 mol L-1 HCl....

A 18.6 mL solution of 0.100 mol L-1 NaOH is titrated using 0.150 mol L-1 HCl. What is the pH of the solution after 15.3 mL of the HCl solution is added?

Calculate the pH of a solution prepared by mixing 20 ml of 9.12 mol/L salicylic acid...

Calculate the pH of a solution prepared by mixing 20 ml of 9.12 mol/L salicylic acid with 20 ml water

In a 0.15M aqueous solution of butanoic acid C3H7CO2H , what is the percentage of butanoic...

In a 0.15M aqueous solution of butanoic acid C3H7CO2H , what is the percentage of butanoic acid that is dissociated? You can find some data that is useful for solving this problem in the ALEKS Data resource. Round your answer to 2 significant digits. Help PLs

Calculate the pH of a solution starting with 200.0 mL of 0.010 M butanoic acid (pKa...

Calculate the pH of a solution starting with 200.0 mL of 0.010 M butanoic acid (pKa = 4.818) solution that has been titrated to the equivalence point with 0.050 M NaOH. Ignore activities and state your answer with 3 sig. figs. Use approximations if you can.

A 360.0 ml buffer solution is 0.130 mol/L in HF and 0.130 mol/L in NaF.

A 360.0 ml buffer solution is 0.130 mol/L in HF and 0.130 mol/L in NaF. Part A : what mass of NaOH could this buffer neutralize before the PH rises above 4.00? Part B: If the same volume of the buffer was 0.370mol/L in NaF, what mass of NaOH could be handled before the pH rises above 4.0? the Ka of HF is 3.5*10^-4 this is all the information given, I can look up more ka or...

Subjects