Chemistry, 22.07.2019 02:30 blakestuhan

A25.0 ml sample of 0.105 m hcl was titrated with 31.5 ml of naoh. what is the concentration of the naoh?

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The balanced equation for the above reaction is
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCk is 11:
number of HCl moles reacted - 0.105 mol/L x 0.0250 L = 0.00263 mol
according to molar ratio, number of HCl moles = number of NaOH moles
therefore number of NaOH moles - 0.00263 mol
number of moles in 31.5 mL - 0.00263 mol
therefore number of NaOH moles in 1000 mL = 0.00263 / 31,5 mL x 1000 mL
molarity of NaOH - 0.0835 M

Answer from: michello3

The concentration of the NaOH: 0.083 M

Further explanation

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint / equivalence point of the reaction can use indicators according to the appropriate pH range

Titrations can be distinguished including acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases is used.

An acid-base titration there will be a change in the pH of the solution.

From this pH change a Titration Curve can be made which is a function of acid / base volume and pH of the solution

Acid-base titration formula

Ma Va. na = Mb. Vb. nb

Ma, Mb = acid base concentration

Va, Vb = acid base volume

na, nb = acid base valence

Neutralization Reaction:

NaOH + HCl ⇒ NaCl + H₂O

25.0 ml, 0.105 M HCl was titrated with 31.5 ml of NaOH

Acid-base titration formula

Ma Va. na = Mb. Vb. nb

a = HCl, b = NaOH (both have valence 1)

0.105 M. 25 = Mb. 31.5. 1

$\rm Mb = \dfrac {0.105 \times 25} {31.5} \\\\ Mb = \boxed {\bold {0.083 \: M}}$

Learn more

the best tool to find the volume of that she needs to neutralize the acid

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Determine the endpoint of the titration.

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acid-base

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Identify the acids and the bases

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endpoint titration

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A25.0 ml sample of 0.105 m hcl was titrated with 31.5 ml of naoh. what is the concentration of the n...

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