Question

1a)What volume (to the nearest 0.1 mL) of 7.10-M HCl must be added to 0.550 L of 0.350-M K2HPO4 to prepare a pH = 7.50 buffer? b)What volume (to the nearest 0.1 mL) of 7.60-M NaOH must be added to 0.700 L of 0.200-M HNO2 to prepare a pH = 3.20 buffer? c)How many mL (to the nearest mL) of 0.250-M KF solution should be added to 740. mL of 0.220-M HF to prepare a pH = 3.10 solution?

b)What volume (to the nearest 0.1 mL) of 7.60-M NaOH must be added to 0.700 L of 0.200-M HNO₂ to prepare a pH = 3.20 buffer?

c)How many mL (to the nearest mL) of 0.250-M KF solution should be added to 740. mL of 0.220-M HF to prepare a pH = 3.10 solution?

Answer 1

ans)

1)

a)

from above data that

the reaction is

HCl      +     HPO₄^2-    ---->       Cl^-    +    H₂PO₄^-

initial:       7.10x          0.1925                     0               0
delta:      -7.10x          -7.10x                   7.10x         7.10x
equil:          0         0.1925 - 7.10x           7.10x         7.10x

we get

0.1925 = 0.350 x 0.550

we know that the equation

: pH = pK_a + log(n_b/n_a)

The p_Ka for HPO₄^2- = 7.21

So substitue the values

7.5 = 7.21 + log([HPO₄^2-]/[H₂PO₄^-])

7.5 = 7.21 + log((0.1925 - 7.10x)/(7.10x))

Solve for x and this will be the amount of HCl in L to prepare a pH = 7.50 buffer.

x = 9.19 10^-3 L = 9.19 mL

b)

from above data that

First set up a reaction table like so:

           HNO2 + OH- <---> NO2- + H2O

You are given the amount of acid in the problem:

0.700L x 0.200M = 0.14 moles HNO2

The question wants to know the initial base (OH-)

We have no initial products and water is neglected for thistable:

           HNO2 +       OH- <---> NO2- + H2O
initial    0.14 x                  0              -
delta         -x           -x                +x             -
final      0.14-x    ~0                  x              -

Use the Henderson-HASSELBALCH equation to determine the ratio ofBase/Acid:

(Note that I have rearranged it and simplified it to finish thisquickly)

10^{(pH - pka)} = Base / Acid

10^(3.20-3.40) = 0.6309 (This is our ratio ofBase/Acid)

Determine moles of Base needed using the ratio wedetermined in step 2

0.6309= Base / Acid

0.6309= (x) / (0.14-x)
(These values came from our reaction table final line)

Solving for x = 0.054157(this is the moles of OH that need to beadded)

Convert moles of OH to mL:

0.054157 moles x 1000 mL / 7.6 moles = 7.13 mL

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