By titration, 15.0 mL of 0.1008 M sodium hydroxide is

kramtus51

kramtus51

Answered question

2022-01-04

By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutra lize a 0.2053-g sample of a weak acid. What is the molar mass of the acid if it is monoprotic?

Answer & Explanation

Ben Owens

Ben Owens

Beginner2022-01-05Added 27 answers

Molarity of NaOH=0.1008M=0.1008molL 
Volume of solution =15mL=0.015L(1mL=0.001L) 
Using the molarity and volume of the solution, we can determine the moles of NaCl.
MNaOH=mo≤s NaOHlitres soln 
Mo≤s NaOH=litres solntimseMNaOH 
=(0.015L soln)(0.1008mol NaOH1L soln) 
=0.001512mol NaOH 
1 mol of NaOH will neutralize 1 mol of H+ in a monoprotic acid. So 0.001512mol NaOH will neutralize 0.001512mol H+. 
1 mol of monoprotic acid contains 1 mol H+. So 0.001512mol H+ is released by 0.001512mol monoprotic acid. 
Moles of monoprotic acid =0.001512mol 
Mass of the monoprotic acid =0.2053g 
Using this equation, we can determine the molar mass.
Mo≤s=MassMolar mass 
Molar mass=MassMo≤s 
=0.2053g0.001512mol 
=135.7gmol

Fasaniu

Fasaniu

Beginner2022-01-06Added 46 answers

Mol of NaOH =15.0mL×(1L1000mL)×(0.1008mol1L)=1.51×10−3mol
mass of acidmol of acid=0.2053g1.51×10−3mol=135.8gmol
xleb123

xleb123

Skilled2023-06-14Added 181 answers

Given information:
Volume of sodium hydroxide (VNaOH) = 15.0 mL
Concentration of sodium hydroxide (CNaOH) = 0.1008 M
Mass of weak acid (macid) = 0.2053 g
We can start by calculating the number of moles of sodium hydroxide used in the neutralization reaction.
nNaOH=CNaOH×VNaOH
Substituting the given values:
nNaOH=(0.1008M)×(15.0mL)
Next, we need to determine the number of moles of the weak acid. Since the weak acid is monoprotic, it reacts with one mole of sodium hydroxide (NaOH) in the neutralization reaction.
nacid=nNaOH
Now we can calculate the molar mass of the acid using the formula:
Molar mass=macidnacid
Substituting the given values:
Molar mass=0.2053gnNaOH
Now, let's substitute the value of nNaOH and calculate the molar mass of the acid.
Molar mass=0.2053g(0.1008M)×(15.0mL)
Simplifying this expression gives us the molar mass of the acid.
fudzisako

fudzisako

Skilled2023-06-14Added 105 answers

To solve the problem, we can use the concept of stoichiometry and the equation for neutralization reactions.
The balanced chemical equation for the neutralization reaction between sodium hydroxide (NaOH) and a monoprotic weak acid (HA) can be written as:
HA+NaOH→NaA+H2O
From the given information, we know that 15.0 mL of 0.1008 M sodium hydroxide solution is needed to neutralize a 0.2053-g sample of the weak acid.
We can start by calculating the number of moles of sodium hydroxide used in the reaction:
moles of NaOH=concentration×volume
moles of NaOH=0.1008M×0.0150L
Next, we need to determine the molar ratio between sodium hydroxide and the weak acid. Since the weak acid is monoprotic, the molar ratio is 1:1. This means that one mole of sodium hydroxide reacts with one mole of the weak acid.
Now, we can calculate the number of moles of the weak acid:
moles of weak acid=moles of NaOH
Given that the molar mass (M) is defined as the mass (in grams) divided by the number of moles, we can rearrange the equation to solve for the molar mass of the weak acid:
M=massmoles of weak acid
Plugging in the values, we have:
M=0.2053gmoles of NaOH
Substituting the moles of NaOH calculated earlier, we get:
M=0.2053g0.1008M×0.0150L

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Other

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?

Product

Company

Connect with us

Get Plainmath App

Plainmath Logo

E-mail us: [email protected]

Our Service is useful for:

Plainmath is a platform aimed to help users to understand how to solve math problems by providing accumulated knowledge on different topics and accessible examples.

2023 Plainmath. All rights reserved