Question #3c224

Question

Question #3c224

1 Answer

Impact of this question

2041 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-03T23:26:58

The answer is $23.0 mL$ $"23.0 mL"$ of disodium phosphate solution is needed for this particular buffer.

So, you know that the total volume of the buffer will be $40.0 mL$ $"40.0 mL"$ , or $0.04 L$ $"0.04 L"$ . This means that the sum of the volumes of the two solutions must be

$V_{1} + V_{2} = 0.04$ $V_1 + V_2 = 0.04$ (1)

SInce you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine another relationship between $V_{1}$ $V_1$ and $V_{2}$ $V_2$

#pH = pKa + log(([Na_2HPO_4])/([NaH_2PO_4]))#

$6.95 = 6.82 + log (\frac{[N a_{2} H P O_{4}]}{[N a H_{2} P O_{4}]}) \Rightarrow \frac{[N a_{2} H P O_{4}]}{[N a H_{2} P O_{4}]} = 1.35$ $6.95 = 6.82 + log(([Na_2HPO_4])/([NaH_2PO_4])) => ([Na_2HPO_4])/([NaH_2PO_4]) = 1.35$ (2)

Now, use the initial concentrations of the two solutions to find the relationship you need. For sodium phosphate, the buffer concentration can be written as

$[N a H_{2} P O_{4}] = \frac{n_{sodium}}{V_{1} + V_{2}}$ $[NaH_2PO_4] = n_("sodium")/(V_1+V_2)$

The number of moles of sodium phosphate is

$n_{sodium} = C \cdot V_{1} = 0.100 M \cdot V_{1} = 0.100 \cdot V_{1}$ $n_("sodium") = C * V_1 = "0.100 M" * V_1 = 0.100 * V_1$

Likewise, the concentration of disodium phophate is

$[N a_{2} H P O_{4}] = \frac{n_{disodium}}{V_{1} + V_{2}}$ $[Na_2HPO_4] = n_("disodium")/(V_1+V_2)$

Once again, the number of moles can be calculated using the initial concentration

$n_{disodium} = C \cdot V_{2} = 0.100 M \cdot V_{2} = 0.100 \cdot V_{2}$ $n_("disodium") = C * V_2 = "0.100 M" * V_2 = 0.100 * V_2$

Plug all of this into equation (2)

$\frac{0.100 \cdot V_{2}}{V_{1} + V_{2}} \cdot \frac{V_{1} + V_{2}}{0.100 \cdot V_{1}} = 1.35$ $(0.100 * V_2)/(V_1 + V_2) * (V_1 + V_2)/(0.100 * V_1) = 1.35$ , or

$\frac{V_{2}}{V_{1}} = 1.35 \Rightarrow V_{2} = 1.35 \cdot V_{1}$ $V_2/V_1 = 1.35 => V_2 = 1.35 * V_1$ . Plug this into equation (1)

$V_{1} + 1.35 \cdot V_{1} = 0.04 \Rightarrow V_{1} = \frac{0.04}{2.35} = 0.017$ $V_1 + 1.35 * V_1 = 0.04 => V_1 = 0.04/2.35 = 0.017$ , which means that

$V_{2} = 0.04 - 0.017 = 0.023$ $V_2 = 0.04 - 0.017 = 0.023$

Therefore, the volume of disodium phosphate $0.100 M$ $"0.100 M"$ solution you need for this particular buffer is

$V_{2} = 0.023 L = 23.0 mL$ $V_2 = "0.023 L"= "23.0 mL"$

Science

Math

Humanities

... and beyond

Question #3c224

1 Answer

Related questions

Impact of this question