What is the $pH$ $"pH"$ for an aqueous solution with a hydroxide concentration of $5 \times 10^{- 3} M$ $5 xx 10^(-3) "M"$ at $25^{\circ} C$ $25^@ "C"$ ?

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What is the $pH$ $"pH"$ for an aqueous solution with a hydroxide concentration of $5 \times 10^{- 3} M$ $5 xx 10^(-3) "M"$ at $25^{\circ} C$ $25^@ "C"$ ?

2 Answers

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Answer 1 · 2017-12-25T02:26:13

$11.7$ $11.7$

Explanation:

$pOH = - log [{OH}^{-}]$ $"pOH" = - log["OH"^-]$

$pOH = - log (5 \times 10^{- 3}) = 3 - log 5 = 2.30$ $"pOH" = - log(5 × 10^-3) = 3 - log5 = 2.30$

Now,

$pH + pOH = 14$ $"pH" + "pOH" = 14$

$pH = 14 - pOH = 14 - 2.30 = 11.7$ $"pH" = 14 - "pOH" = 14 - 2.30 = 11.7$

Answer 2 · 2017-12-26T08:13:02

$pH \approx 11.7$ $"pH" ~~ color(blue)(11.7)$

Explanation:

$p H = - log [H_{3} O^{+}]$ $color (indigo)(pH=-log [H_3O^+])$
$p O H = - log [O H^{-}]$ $color (indigo)(pOH=-log [OH^-])$
$[O H^{-}] = 0.005$ $color(cyan)([OH^-]=0.005)$
Here,
$p O H = - log [0.005]$ $pOH=-log [0.005]$
$p O H = - log [5 \times 10^{- 3}]$ $pOH=-log [5×10^-3]$
$p O H = 2.3010299957$ $pOH=2.3010299957$

$p H = 14 - p O H$ $color (green)(pH=14-pOH)$

$p h = 14 - 2.301 = 11.699$ $ph=14-2.301 =11.699$ $\approx$ $~~$ $11.7$ $bb11.7$

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What is the $pH$ $"pH"$ for an aqueous solution with a hydroxide concentration of $5 \times 10^{- 3} M$ $5 xx 10^(-3) "M"$ at $25^{\circ} C$ $25^@ "C"$ ?

2 Answers

Explanation:

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What is the pH"pH" for an aqueous solution with a hydroxide concentration of 5×10−3M5 xx 10^(-3) "M" at 25∘C25^@ "C"?

2 Answers

Explanation:

Explanation:

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What is the $pH$ $"pH"$ for an aqueous solution with a hydroxide concentration of $5 \times 10^{- 3} M$ $5 xx 10^(-3) "M"$ at $25^{\circ} C$ $25^@ "C"$ ?