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Alla [95]
3 years ago
8

Hclo is a weak acid (ka=4.0x10^-8) and so the salt naclo acts as a weak base. what is the ph of a solution that is 0.037 m in na

clo at 25 c
Chemistry
1 answer:
ankoles [38]3 years ago
7 0

<u>Given:</u>

Concentration of NaClO = 0.037 M

ka (HClO) = 4.0*10⁻⁸

<u>To determine:</u>

The pH of the NaClO solution

<u>Explanation:</u>

The hydrolysis of the weak base can be represented by the ICE table shown below-

                ClO-      +     H2O    ↔    HClO    +       OH-

Initial        0.037M                              0                    0

Change         -x                                 +x                  +x

Equilibrium  (0.037-x)                        x                     x

kb  = kw/ka =  [HClO][OH-]/[ClO-]

10⁻¹⁴/4*10⁻⁸ = x²/(0.037-x)

x = [OH-] = 9.62*10⁻⁵

p[OH-] = -log[OH-] = -log [9.62*10⁻⁵] = 4.02

pH = 14-p[OH-] = 14 - 4.02 = 9.98

Ans: pH of the solution is 9.98


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6 0
2 years ago
HELP - a solution NaOCl was prepared at pH 10.3. calculate the concentration of the salt formed. Given Ka for HOCl=3.0x10^-8
GenaCL600 [577]
First we dissociate the salt

NaOCl ⇒ Na⁺ + OCl⁻
note that [NaOCl] = [OCl⁻]

note that Na⁺ does not undergo hydrolysis so it is a spectator.
OCl⁻ can reform HOCl in an equilibrium, with OCl⁻ acting as the base.

OCl⁻ + H₂O ⇄ OH⁻ + HOCL

<span>Ka for HOCl = 3.0x10^-8, therefore

Kb for OCL</span>⁻ = Kw / (Ka for HOCl)
                       = 1.0 × 10⁻¹⁴ / 3.0 <span>× 10⁻⁸
                       = 3.3 × 10⁻⁷

</span>since Kb for OCl⁻ is given, and the pH is given, then [OCl⁻] must be found, and [OCl⁻] = [NaOCl]

convert the pH into [OH⁻]:
pOH = 14 - pH = 14 - 10.3 = 3.7
[OH⁻] = 10^(-3.7) = 1.995 × 10⁻⁴

set up equib table

          OCl⁻                + H₂O    ⇄      OH⁻        +           HOCl
  ST     x                                               0                             0
 +Δ    -1.995 × 10⁻⁴                   +1.995 × 10⁻⁴        +1.995 × 10⁻⁴
---------------------------------------------------------------------------------
 EQ:   x -1.995 × 10⁻⁴                 1.995 × 10⁻⁴          1.995 × 10⁻⁴

Kb = [OH⁻][HOCl] / [OCl⁻]
3.3 × 10⁻⁷ = (1.995 × 10⁻⁴)² / (x -1.995 × 10⁻⁴ )
x = 1.1963 × 10⁻¹ M = [OCl⁻] = [NaOCl]

The concentration of the salt formed, <span>NaOCl,</span> is 1.2 × 10⁻¹ M

7 0
3 years ago
Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas
Butoxors [25]
<span>Answer: 1.06 moles of molecules = 6.38x10²³ molecules
</span><span />

<span>Explanation:
</span><span />

<span>1) Balanced chemical equation:
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<span>2Na + 2H₂O -> 2NaOH + H₂
</span><span />

<span>2) Mole ratios:
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<span>2 mol Na : 2 mol H₂O : 2 mol NaOH : 1 mol H₂
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<span>3) Convert 48.7 g of sodium to moles:
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<span>number of moles = mass in grams / molar mass
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<span>molar mass of Na = atomic mass = 23.0 g/mol


</span><span>number of moles = 48.7g / 23.0 g/mol = 2.12 moles of Na</span>
<span /><span>
4) Proportion:
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<span>2moles Na/1mol H₂ = 2.12 moles Na / x
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<span>x = 2.12 moles Na * 1 mol H₂ / 2 moles Na = 1.06 moles H₂.
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<span>5) If you want to convert that to number of molecules you have to multiply by Avogadro's number: 6.022 x 10^²³
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<span>1.06 mol x 6.022x10^²³ molecules/mol = 6.38x10²³ molecules</span>
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