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liberstina [14]
2 years ago
6

Calculate the concentration of H3O in a solution that contains 5.5 × 10-5 M OH at 25°C. Identify the solution as acidic, basic o

r neutral. Group of answer choices 5.5 × 10-10 M, neutral 9.2 × 10-1 M, basic 9.2 × 10-1 M, acidic 1.8 × 10-10 M, basic
Chemistry
1 answer:
Masteriza [31]2 years ago
4 0

Answer:

The correct option is option (D).

Therefore the concentration of H_3O is 1.8×10⁻¹⁰ M and it is basic in nature.

Explanation:

List of pH:

  1. If pH of a solution is 7, then solution is neutral.
  2. If pH of a solution is grater than 7, then solution is basic.
  3. If pH of a solution is less than 7, then solution is acidic.

pH of a solution is = - log₁₀[H₃O⁺]

We know that,

[H₃O⁺][OH⁻] = K_w = 10^{-14}  (at 25°C)

Taking log both sides

log ([H₃O⁺][OH⁻]) =log K_w =log 10^{-14}

⇒log[H₃O⁺]+log[OH⁻] = logK_w = -14 log 10

[since log(mn)= log m + log n, log( a^b)=b\ log (a) ]

⇒ - log[H₃O⁺] - log[OH⁻] = - (-14)  [ log 10 =1]

⇒pH+pOH =14.

Given that,

The concentration of OH⁻ is 5.5 × 10⁻⁵ M

[H₃O⁺][OH⁻] = K_w = 10^{-14}

⇒[H₃O⁺] 5.5 × 10⁻⁵ M=10⁻¹⁴

\Rightarrow [H_3O^+]=\frac{10^{-14}}{5.5\times 10^{-5}}

⇒[H₃O⁺] = 1.8×10⁻¹⁰ M

Now check the pH of the solution.

[H₃O⁺] = 1.8×10⁻¹⁰

Taking log both sides

log [H₃O⁺] =log( 1.8×10⁻¹⁰)

⇒ -log [H₃O⁺] = - log( 1.8×10⁻¹⁰)

⇒ pH = - log( 1.8×10⁻¹⁰)

⇒ pH =9.7.

So the nature of solution is basic.

Therefore the concentration of H_3O is 1.8×10⁻¹⁰ M and it is basic in nature.

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Answer:

a) 11.64 M

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c) 1.7 kg

Explanation:

a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:

1.18*1000 = 1180 g.

The mass of HCl will be then:

mHCl = 1180*0.36 = 424.8 g

The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:

nHCl = 424.8/36.5 = 11.64 mol

The molarity is the number of moles divided by the volume in L:

Molarity = 11.64 M

b) To prepare a solution by dilution of a concentrated one, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:

11.64*V1 = 2.00*0.250

V1 = 0.0429 L ≅ 43 mL

c) The neutralization will happen by the equation:

HCl + NaHCO₃ → NaCl + CO₂ + H₂O

So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:

nHCl = 1.75*11.64 = 20.37 mol

The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:

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6 0
2 years ago
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2 years ago
Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because
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Answer:

0.3023 M

Explanation:

Let Picric acid = H_{picric}

So,  H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

The ICE table can be given as:

                          H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

Initial:                0.52                                               0                  0

Change:             - x                                                 + x                 + x

Equilibrium:      0.52 - x                                        + x                 + x

Given that;

acid dissociation constant  (K_a) = 0.42

K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}

0.42 = \frac{[x][x]}{0.52-x}}

0.42 = \frac{[x]^2}{0.52-x}}

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x²  + 0.42x - 0.2184 = 0                   -------------------- (quadratic equation)

Using the quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}    ;     ( where +/-  represent ± )

= \frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}

= \frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}

= \frac{-0.42+\sqrt {1.0496} }{2}     OR   \frac{-0.42-\sqrt {1.0496} }{2}

= \frac{-0.42+1.0245}{2}       OR    \frac{-0.42-1.0245}{2}

= \frac{0.6045}{2}                 OR    -\frac{1.4445}{2}

= 0.30225          OR     - 0.72225

So, we go by the +ve integer that says:

x =  0.30225

x = [ H_3}O^+ ] = [   Picric^- ] =  0.3023  M

∴  the value of  [H3O+] for an 0.52 M solution of picric acid  = 0.3023 M     (to 4 decimal places).

6 0
2 years ago
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Answer:

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Amontons’s law: P/T = Constant at constant V and n

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Also temperature T2 of 40°C = 313.15 K

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