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meriva
3 years ago
15

A form of energy that is stored in a stretched rubber band is called

Chemistry
1 answer:
storchak [24]3 years ago
6 0
You input potential (stored<span>) </span>energy<span> into the </span>rubber band<span> system when you </span>stretched<span> the</span>rubber band<span> back. Because it is an </span>elastic<span> system, this kind of potential </span>energy<span> is specifically </span>called elastic<span> potential </span>energy<span>. ... When the </span>rubber band<span> is released, the potential </span>energy<span> is quickly converted to kinetic (motion) </span>energy<span>.</span>
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The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

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