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Fittoniya [83]
3 years ago
6

: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be

the new concentration of HC6H5O7 2- after 25.0 mL of 0.125 M NaOH is added to the buffer?
Chemistry
1 answer:
Jet001 [13]3 years ago
4 0

Answer: New concentration of HC_{6}H_{5}O^{2-}_{7} is 0.23 M.

Explanation:

The given data is as follows.

     Moles of HC_{6}H_{5}O^{2-}_{7} = 0.06 mol

     Moles of H_{2}C_{6}H_{5}O_{7} = 0.08 mol

Therefore, moles of OH^{-} added are as follows.

    Moles of OH^{-} = 0.125 \times \frac{25}{1000}

                          = 0.003125 mol

Now, new moles of HC_{6}H_{5}O^{2-}_{7} = 0.06 + 0.003125

                     = 0.063125

Therefore, new concentration of HC_{6}H_{5}O^{2-}_{7} will be calculated as follows.

       Concentration = \frac{0.063125}{0.275}

                               = 0.23 M

Thus, we can conclude that new concentration of HC_{6}H_{5}O^{2-}_{7} is 0.23 M.

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4 0
3 years ago
How many molecules are present in 265 moles of CCl
Solnce55 [7]

Answer:

Option C. 1.60x10^26 molecules

Explanation:

Avogadro's hypothesis proved that that 1 mole of any substance contains 6.02x10^23 molecules.

From the above, we understood that 1 mole of CCl4 contains 6.02x10^23 molecules.

If 1 mole of CCl4 contains 6.02x10^23 molecules,

then, 265 moles of CCl4 will contain = 265 x 6.02x10^23 = 1.60x10^26 molecules

From the calculation made above, 265 moles of CCl4 contains 1.60x10^26 molecules.

5 0
3 years ago
what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?
MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now  F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻  +  H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF,  the [ OH ⁻] from the given pH, and finally the mass needed to produce that  OH⁻ concentration.  

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is  6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14  ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

                                   F⁻                    HF                        OH⁻

Equilibrium                 X                  2.51 x 10⁻⁶            2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X  =  1.5 x 10⁻¹¹     ⇒  X =  (2.51 x 10⁻⁶)²  / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L  *  0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF *  42.0 g/mol NaF = 6.03 g NaF

7 0
3 years ago
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