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3 years ago

Write the balcanced molecular and net ionic equation for the acid-base reaction:

Chemistry

1 answer:

poizon [28]3 years ago

7 0

Explanation: For the given acid-base reaction, The molecular equation is:

Molecular equation: $HNO_3(aq.)+CH_3NH_2(aq.)\rightarrow CH_3NH_3^+(aq.)+NO_3^-(aq.)$

As we know, $HNO_3$ is a strong acid, so in the ionic equation, it will completely dissociate into its respective ions.

The ionic equation for the given acid-base reaction is given by:

$H^+(aq.)+NO_3^-(aq.)+CH_3NH_2(aq.)\rightarrow CH_3NH_3^+(aq.)+NO_3^-(aq.)$

As nitrate ions are present on both the sides, so in the net ionic equation, it will get cancelled out, then the net ionic equation becomes:

Net ionic equation: $H^+(aq.)+CH_3NH_2(aq.)\rightarrow CH_3NH_3^+(aq.)$

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4 years ago

A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increa

MatroZZZ [7]

Answer:

T2= 7.3°C

Explanation:

To solve this problem we will use Charles law equation i.e,

V1/T1 = V2/T2

Given data

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3 0

3 years ago

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,

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3 years ago

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3 years ago