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BaLLatris [955]
3 years ago
14

if it takes 54 mL of 0.1 NaOH to neutralize 125 mL of an HCL solution, what is the concentration of HCL?

Chemistry
1 answer:
OleMash [197]3 years ago
3 0

Answer:

0.0432M

Explanation:

We begin by writing a balanced equation for the reaction. This is illustrated below:

NaOH + HCl —> NaCl + H2O

From the equation above,

The number of mole of the acid (nA) = 1

The number of mole of the base (nB) = 1

Data obtained from the question include:

Vb (volume of the base) = 54mL

Cb (concentration of the base) = 0.1M

Va (volume of the acid) = 125mL

Ca ( concentration of the acid) =?

Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:

CaVa/CbVb = nA/nB

Ca x 125 / 0.1 x 54 = 1

Cross multiply to express in linear form:

Ca x 125 = 0.1 x 54

Divide both side by 125

Ca = (0.1 x 54) / 125

Ca = 0.0432M

Therefore, the concentration of the acid is 0.0432M

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alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decima
svetlana [45]

A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = \mathtt{ \dfrac{0.100  \ mol}{ 1.0  \  L }}  = 0.10 M

The chemical equation for this reaction is :

\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = \mathtt{ \dfrac{0.100  \ mol}{ 1.0  \  L }}  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = \mathtt{ \dfrac{0.035  \ mol}{ 1.0  \  L }} = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  \mathtt{CH_3COOH \ \ \   +  \ \ \ \ OH^-  \ \ \to \ \ CH_3COO^-  \ \ \ + \ \ \  H_2O}

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log \mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}

The pH of this solution = 4.74 + log \mathtt{\dfrac{0.135}{0.065}}

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places

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