A solution of NaOH needed to be standardized before use in other titrations. If 0.356 g...
A solution of NaOH needed to be standardized before use in other titrations. If 0.356 g of KHP (MW = 204.23 g/mol ) required 30.89 mL of the base to reach the equivalence point, what is the concentration of NaOH? Write a balanced equation to verify titrant: analyte ratio.
Homework Answers
moles = mass / molar mass
so
moles of KHP = 0.356 g / (204.23 g/mol) = 1.743 x 10-3 mol
now
KHP + NaOH ==> KNaP + H2O
so
moles of NaOH required = 1.743 x 10-3 mol KHP x (1 mol NaOH / 1 mol KHP)
moles of NaOH required = 1.743 x 10-3 mol NaOH
now
concentration = moles of NaOH / volume of base required (L)
concentration of NaOH = 1.743 x 10-3 mol / 0.03089 L
concentration of NaOH = 0.05643 mol / L
concentration of NaOH = 0.05643 M
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A solution of NaOH needed to be standardized before use in other
titrations. If 0.356 g...
I MAINLY NEED HELP ON TABLES PLS HELP THANK YOU Materials: NaOH MW = 40 g/mL,...
I MAINLY NEED HELP ON TABLES
PLS HELP THANK YOU
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC2H4O4, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 mL water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the...
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PRE LAB : Volumetric Titrations. (Acid-Base Titrations) Name: ID Date 1. How many mL of a 0.103M NaOH solution are required to neutralize 10.00mL of a 0.198M HCI solution? 2. what is the difference between end point and equivalence point? 3. A titration is performed and 20.70 mL of 0.500M KOH is required to reach the end point when titrated against 15.00 mL of H2SO4 of unknown concentration. Write the chemical equation and solve for the molarity of the acid....
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol....
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol....
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
Question 4 You mass out 0.554 g of an impure sample of KHP. You use 17.25...
Question 4 You mass out 0.554 g of an impure sample of KHP. You use 17.25 mL of your standardized 0.0994 M NaOH to reach the end point. What percentage of KHP is in your impure sample? The MW of KHP is 204.23 g/mol. The MW of NaOH is 40.00 g/mol. Group of answer choices 63.2 % KHP 35.018 % KHP 12.4 % KHP 35.0 % KHP
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Molarity for NaOH = 0.08732. In a second titration with the same solution of NaOH as used in Question #1, the student weighs out a sample of KHP of 0.359 g. Calculate the volume of the NaOH solution needed to neutralize this sample of KHP. 3. A monoprotic weak acid with the general formula of HA will react with a base, such as NaOH. Write the neutralization equation which describes the reaction. 4. If K, for the weak acid, HA is 1.8...
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Sodium hydroxide is standardized by a reaction with potassium hydrogen phtalate (KHP, MW 2042 g/mol). Calculate the molarity of a NaOH solution if 0.5123 g KHP were titrated to the equivalence point with 22.75 mL solution of NaOH. 0.0002509 moll Ob 0.02275 moll O c.0.1103 mol/l Od 1.10 mol/l
A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.2029 M solution of NaOH at 25° C. (a) What is...
A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.2029 M solution of NaOH at 25° C. (a) What is the pH of the HF solution before titrant is added? (b) How many milliliters of titrant are required to reach the equivalence point? mL (c) What is the pH at 0.50 mL before the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH at 0.50 mL after the equivalence point?
Part A: Standardization of NaOH Solution Data Table: Trial 1 Trial 2 Trial 3 Tared mass...
Part A: Standardization of NaOH Solution Data Table: Trial 1 Trial 2 Trial 3 Tared mass of KHP (g) 5065 -S168 so40 Molar mass of KHP (g/mol) 204.23 204.23 204.23 Moles of KHP (mol) Buret Reading:Initial Volume of NaOH (mL) .3 23. 2 나6.2. Buret Reading: Final Volume of NaOH (mL) Volume of NaOH dispensed (mL) Molar concentration of NaOH (mol/L) Average Molar concentration of NaOH 28.3 22.1 1145 O533 ces 3 (mol/L) Experiment: Volumetric Analvsis
1. A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known...
1. A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titrated with the NaOH solution until a light pink color appeared using phenolpthalein indicator. Using the volume of NaOH required to neutralize KHP and the number of moles of KHP titrated, the concentration of the NaOH solution was calculated. Molecular formula of Potassium hydrogen phthalate: HKC8H404 Mass of KHP used for standardization (g) 0.5100 Volume of NaOH required to neutralize...
I MAINLY NEED HELP ON TABLES
PLS HELP THANK YOU
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC2H4O4, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 mL water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the...
PRE LAB : Volumetric Titrations. (Acid-Base Titrations) Name: ID Date 1. How many mL of a 0.103M NaOH solution are required to neutralize 10.00mL of a 0.198M HCI solution? 2. what is the difference between end point and equivalence point? 3. A titration is performed and 20.70 mL of 0.500M KOH is required to reach the end point when titrated against 15.00 mL of H2SO4 of unknown concentration. Write the chemical equation and solve for the molarity of the acid....
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
Materials: NaOH MW = 40 g/mL, KHP - potassium hydrogen phthalate, KHC8H404, MW = 204.23 g/mol. Acetic acid, HC2H302, MW = 60.05 g/mol Part 1: Standardization of NaOH Assume 0.951 g of KHP is weighed and transferred to a 250 mL Erlenmeyer flask. Approximately 50 ml water is added to dissolve the KHP. Note that the exact volume of water is not important because you only need to know the exact number of moles of KHP that will react with...
Molarity for NaOH = 0.08732. In a second titration with the same solution of NaOH as used in Question #1, the student weighs out a sample of KHP of 0.359 g. Calculate the volume of the NaOH solution needed to neutralize this sample of KHP. 3. A monoprotic weak acid with the general formula of HA will react with a base, such as NaOH. Write the neutralization equation which describes the reaction. 4. If K, for the weak acid, HA is 1.8...
Sodium hydroxide is standardized by a reaction with potassium hydrogen phtalate (KHP, MW 2042 g/mol). Calculate the molarity of a NaOH solution if 0.5123 g KHP were titrated to the equivalence point with 22.75 mL solution of NaOH. 0.0002509 moll Ob 0.02275 moll O c.0.1103 mol/l Od 1.10 mol/l
A 35.00-mL solution of 0.2500 M HF is titrated with a standardized 0.2029 M solution of NaOH at 25° C. (a) What is the pH of the HF solution before titrant is added? (b) How many milliliters of titrant are required to reach the equivalence point? mL (c) What is the pH at 0.50 mL before the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH at 0.50 mL after the equivalence point?
Part A: Standardization of NaOH Solution Data Table: Trial 1 Trial 2 Trial 3 Tared mass of KHP (g) 5065 -S168 so40 Molar mass of KHP (g/mol) 204.23 204.23 204.23 Moles of KHP (mol) Buret Reading:Initial Volume of NaOH (mL) .3 23. 2 나6.2. Buret Reading: Final Volume of NaOH (mL) Volume of NaOH dispensed (mL) Molar concentration of NaOH (mol/L) Average Molar concentration of NaOH 28.3 22.1 1145 O533 ces 3 (mol/L) Experiment: Volumetric Analvsis
1. A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titrated with the NaOH solution until a light pink color appeared using phenolpthalein indicator. Using the volume of NaOH required to neutralize KHP and the number of moles of KHP titrated, the concentration of the NaOH solution was calculated. Molecular formula of Potassium hydrogen phthalate: HKC8H404 Mass of KHP used for standardization (g) 0.5100 Volume of NaOH required to neutralize...
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