Question #6d0d9

Question

Question #6d0d9

1 Answer

Impact of this question

1068 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-06T19:15:14

A: 50 ml
B:pH=2.87
C:pH=4.74
D:pH=8.72

Explanation:

A:
50 ml HAc 0.1M = 5mMol HAc (weak acid)
50 NaOH 0.1 M = 5 mMol NaOH (strong base)
Equivalence: 1:1, 5mMol:5 mMol => 50ml 0,.1 M NaOH

B:
$p H = - log \sqrt{K a \cdot F}$ $pH = -log sqrt(Ka * F)$
=> $p H = - log \sqrt{{1.8}^{- 5} * 0.1 M}$ $pH = -log sqrt(1.8^-5 **0.1M)$
=> $p H = - log \sqrt{{1.8}^{6}}$ $pH = -log sqrt(1.8^6)$
=> $p H = - log 1.34 10^{- 3}$ $pH= -log 1.34 10^-3$
=> $p H = 2.87$ $pH = 2.87$

C:
$p H = p K a + log (\frac{η O H^{-}}{η H A_{init} - η O H^{-}})$ $pH = pKa + log((eta OH^-)/(eta HA_"init" - eta OH^-))$

$p H = p K a + log (\frac{{2.5}^{- 3}}{{5.0}^{- 3} - {2.5}^{- 3}})$ $pH = pKa + log(2.5 ^-3/(5.0^-3 - 2.5 ^-3))$

=> $p H = p K a + log (\frac{{2.5}^{- 3}}{{2.5}^{- 3}})$ $pH = pKa + log(2.5^-3/2.5^-3)$
=> $p H = p K a + log 1$ $pH = pKa + log1$
$log 1$ $log1$ = 0, so:
pH=pKa = $- log K a = 4.74$ $-logKa = 4.74$

D:
$p H = 14 + log \sqrt{\frac{C_{a} \cdot C_{b} \cdot K_{w}}{(C_{a} + C_{b}) \cdot K a}}$ $pH= 14 + log sqrt((C_a* C_b* K_w) / ((C_a + C_b)* Ka))$

$p H = 14 + log \sqrt{\frac{0.1 \cdot 0.1 \cdot 1 \cdot 10^{- 14}}{(0.1 + 0.1) \cdot 1.8 \cdot 10^{- 5}}}$ $pH = 14 + log sqrt((0.1 * 0.1 * 1*10^-14) / ((0.1 + 0.1 )* 1.8*10^-5))$
$p H = 14 + log \sqrt{\frac{1 \cdot 10^{- 16}}{0.2 \cdot 1.8 \cdot 10^{- 5}}}$ $pH = 14 + log sqrt ((1*10^-16)/(0.2 * 1.8*10^-5))$
$p H = 14 + log \sqrt{\frac{1 \cdot 10^{- 10}}{3.6}} = 14 - 5.28$ $pH = 14 + log sqrt ((1*10^-10)/(3.6)) = 14 -5.28$
$\Rightarrow p H = 8.72$ $=> pH = 8.72$

Science

Math

Humanities

... and beyond

Question #6d0d9

1 Answer

Explanation:

Related questions

Impact of this question