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2 years ago

What is the molarity of an h3po4 solution if 15.0 ml is completely neutralized by 38.5 ml of 0.150m naoh?

Chemistry

2 answers:

polet [3.4K]2 years ago

6 0

The correct answer is 0.128M H3PO4

just olya [345]2 years ago

4 0

The molarity of H3PO4 solution if 15.0 ml is completely neutralized by 38.5 ml of 0.15m naoh is calculated as follows
find moles of NaOH used = molarity x volume
= 38.5 x 0.15 = 5.775 moles

write the reacting equation
3NaOH + H3PO4 = Na3PO4 + 3H2O

from the equation the reacting ratio between NaOH to H3Po4 which is 3:1 the moles of H3PO4 is therefore = 5.775/3 = 1.925 moles

molarity of H3PO4 is therefore = moles /volume
= 1.925/15 = 0.128 M

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At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)

viktelen [127]

Answer: $K_c$ for this reaction at this temperature is 0.029

Explanation:

Moles of $HBr$ = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of $HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M$

The given balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial conc. 0.500 M 0 M 0 M

At eqm. conc. (0.500-2x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}$

Equilibrium concentration of $[Br_2]$ = x = 0.0955 M

Now put all the given values in this expression, we get :

$K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}$

$K_c=0.029$

Thus $K_c$ for this reaction at this temperature is 0.029

7 0

2 years ago

A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid solution to the stoi- chiometric point.

Dmitrij [34]

Answer:

Molar concentration of the weak acid solution is 0.0932

Explanation:

Using the formula: $\frac{C_aV_a}{C_bV+b} = \frac{n_a}{n_b}$

Where Ca = molarity of acid

Cb = molarity of base = 0.0981 M

Va = volume of acid = 25.0 mL

Vb = volume of base = 23.74 mL

na = mole of acid

nb = mole of base

Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1

Therefore, $C_a = \frac{C_bV_b}{V_a}$

Ca = 0.0981 x 23.74/25.0

= 0.093155 M

To 4 significant figure = 0.0932 M

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1 year ago

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densk [106]

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