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tensa zangetsu [6.8K]
3 years ago
9

Calculate the molarity of an hcl solution if a 10.00 ml sample requires 25.24 ml of a 1.600 m naoh solution to be neutralized.

Chemistry
1 answer:
AlladinOne [14]3 years ago
6 0

Answer:

The molarity of the HCl solution should be 4.04 M

Explanation:

<u>Step 1:</u> Data given

volume of HCl solution = 10.00 mL = 0.01 L

volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L

<u>Step 2:</u> The balanced equation

HCl + NaOH → NaCL + H2O

Step 3: Calculate molarity of HCl

n1*C1*V1 = n2*C2*V2

Since the mole ratio for HCl and NaOH is 1:1 we can just write:

C1*V1 =C2*V2

⇒ with C1 : the molarity of HCl = TO BE DETERMINED

⇒ with V1 = the volume og HCl = 10 mL = 0.01 L

⇒ with C2 = The molarity of NaOH = 1.6 M

⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L

C1 * 0.01 = 1.6 * 0.02524

C1 = (1.6*0.02524)/0.01

C1 = 4.04M

The molarity of the HCl solution should be 4.04 M

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olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

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3 years ago
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Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

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2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

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Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

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