How are journals helpful to scientists

what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require

vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

$HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O$

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of $HCOO^{-}$

So, moles of NaOH used to reach equivalence point equal to number of moles $HCOO^{-}$ produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of $HCOO^{-}$ produced = $\frac{29.80\times 0.3567}{1000}moles=0.01063moles$

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of $HCOO^{-}$ = $\frac{0.01063\times 1000}{54.80}M=0.1940M$

At equivalence point, pH depends upon hydrolysis of $HCOO^{-}$ . So, we have to construct an ICE table.

$HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}$

I: 0.1940 0 0

C: -x +x +x

E: 0.1940-x x x

So, $\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}$

or, $x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0$

So, $x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}$

So, $x=3.285\times 10^{-6}M$

So, $pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52$

5 0

2 years ago

What is 25 milliliters in cubic centimeters

lubasha [3.4K]

1cc (cubic centimeter) and 1mL (milliliter) are the same volume. So, 25mL = 25cc

3 0

3 years ago

If O2 molecule is saturated with outer electrons how can it gain electrons to form ions?

PIT_PIT [208]

Answer: it will gain them by multiplying the electrons

Explanation:

its like 2 + 2=4

4+4=8

8 0

1 year ago

How many bananas are equal to 7.50 moles of bananas?

Veseljchak [2.6K]

Answer:

4.52×10^24

Explanation:

N = n × Na

where; N = no. of bananas

n = no. of moles

Na = Avogadro's constant

Which is 6.02×10^23

N = 7.5 × 6.02×10^23

N =4.515×10^24

7 0

2 years ago

Consider the chemical equations shown here.

vichka [17]

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Explanation:

The overall equation for the reaction that produces P₄0₁₀ is :

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Now let us derive this equation:

Given equations:

P₄ $_{s}$ + 30₂ $_{g}$ ⇒ P₄0₆ $_{s}$ equation 1;

P₄ $_{s}$ + 50₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$ equation 2;

To get the overall combined equation, the equation 1 must be reversed and added to equation 2:

P₄0₆ $_{s}$ ⇒ P₄ $_{s}$ + 30₂ $_{g}$ equation 3

+

equation 2:

P₄ $_{s}$ + 50₂ $_{g}$ + P₄0₆ $_{s}$ ⇒ P₄0₁₀ $_{s}$ + P₄ $_{s}$ + 30₂ $_{g}$

cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

learn more:

Net equation brainly.com/question/2947744

#learnwithBrainly

5 0

2 years ago