Question

A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water...

A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water to make a 20.0 mL solution and then this solution was titrated by 0.500 mol/L HCl solution. If 16.1 mL of HCl was used to reach the end point, what is the mass % of NaOH in the sample?

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water...
A sample contains both NaOH and NaCl. 0.500 g of this sample was dissolved in water to make a 20.0 mL solution and then this solution was titrated by 0.500 mol/L HCl solution. If 18.3 mL of HCl was used to reach the end point, what is the mass % of NaOH in the sample? NaOH + HCl --> NaCl + H2O Note: 1. Keep 3 sig figs for your final answer. 2. Answer as percentage.
HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq) 4.00 mL of an unknown...
HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq) 4.00 mL of an unknown acid solution containing HCl was added to flask containing 30.00 mL of deionized water and two drops of an indicator. The solution was mixed and then titrated with NaOH until the end point was reached. Use the following titration data to determine the mass percent of HCl in the acid solution. Mass of flask: 125.59 g Mass of flask + acid solution: 129.50 g...
What is the strength (in ppm) of 0.500 g NaCl dissolved in 1.06 L of water...
What is the strength (in ppm) of 0.500 g NaCl dissolved in 1.06 L of water (Density = 1.000 kg/L). Answer in both % and ppm (mg/L).
A 47.50 mL aliquot from a 0.455 L solution that contains 0.500 g of MnSO4 (Fw...
A 47.50 mL aliquot from a 0.455 L solution that contains 0.500 g of MnSO4 (Fw = 151.00 g/mol) required 36.3 mL of an EDTA solution to reach the end point in a titration. What mass (in milligrams) of CaCO3 (FW = 100.09 g/mol) will react with 1.33 mL of the EDTA solution?
a. A 0.6740-g sample of impure Ca(OH)2 is dissolved in enough water to make 57.70 mL...
a. A 0.6740-g sample of impure Ca(OH)2 is dissolved in enough water to make 57.70 mL of solution. 20.00 mL of the resulting solution is then titrated with 0.2546-M HCl. What is the percent purity of the calcium hydroxide if the titration requires 17.83 mL of the acid to reach the endpoint? b. What is the iodide ion concentration in a solution if the addition of an excess of 0.100 M Pb(NO3)2 to 33.8 mL of the solution produces 565.2...
2. A student dissolved 2.636 g of a powder containing unknown masses of NaCl and KBr...
2. A student dissolved 2.636 g of a powder containing unknown masses of NaCl and KBr in 30.00 mL of de-ionized water and titrated the solution with 0.0500 M AgNO3. If it takes 33.62 mL of Ag the Fajan’s end point, what is the mass of KBr in the powder?   
A 1.762 g sample of impure H2C2O4 (MW =90.04) was dissolved in water and the impurities...
A 1.762 g sample of impure H2C2O4 (MW =90.04) was dissolved in water and the impurities removed by filtration. 50.00 mL of .01028 M NaOH was added to the filtrate, this being in excess of the amount needed to neutralize both hydrogens of the H2C2O4. The excess NaOH required 12.62 mL of 0.1251M HCl to reach the equivalence point. What was the weight % of H2C2O4 in the sample.
A 1.762 g sample of impure H2C2O4 (MW =90.04) was dissolved in water and the impurities...
A 1.762 g sample of impure H2C2O4 (MW =90.04) was dissolved in water and the impurities removed by filtration. 50.00 mL of .01028 M NaOH was added to the filtrate, this being in excess of the amount needed to neutralize both hydrogens of the H2C2O4. The excess NaOH required 12.62 mL of 0.1251M HCl to reach the equivalence point. What was the weight % of H2C2O4 in the sample.
A mixture contains only NaCl and Al2(SO4)3. A 1.75-g sample of the mixture is dissolved in...
A mixture contains only NaCl and Al2(SO4)3. A 1.75-g sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of Al(OH)3. The precipitate is filtered, dried, and weighed. The mass of the precipitate is 0.134 g. What is the mass percent of Al2(SO4)3 in the sample?
A 0.115 g sample of a diprotic acid of unknown molar mass is dissolved in water...
A 0.115 g sample of a diprotic acid of unknown molar mass is dissolved in water and titrated with 0.1208 M NaOH. The equivalence point is reached after adding 14.8 mL of base. What is the molar mass of the unknown acid?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT