Calculating pH

Chemistry Level 2

What is the pH \text{pH} of a neutral solution at 3 7 o C 37^{o}\text{C} , where K w K_{w} equal 2.5 × 1 0 − 14 2.5 \times 10^{-14} ?

6.6 6.8 13.6 7.0

Kushal Patankar by Kushal Patankar , 22, India

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1 solution

Maximos Stratis
Jun 4, 2017

In a neutral solution it is true that:
[ H 3 O + ] = [ O H − ] [H_{3}O^{+}]=[OH^{-}]
Also:
K w = [ H 3 O + ] ⋅ [ O H − ] ⇒ K_{w}=[H_{3}O^{+}]\cdot [OH^{-}]\Rightarrow
[ H 3 O + ] 2 = 2.5 ⋅ 1 0 − 14 ⇒ [H_{3}O^{+}]^{2}=2.5\cdot 10^{-14}\Rightarrow
[ H 3 O + ] = 2.5 ⋅ 1 0 − 7 [H_{3}O^{+}]=\sqrt{2.5}\cdot 10^{-7}
But p H = − l o g ( [ H 3 O + ] ) ⇒ pH=-log([H_{3}O^{+}])\Rightarrow
p H = − l o g ( 2.5 ⋅ 1 0 − 7 ) ⇒ pH=-log(\sqrt{2.5}\cdot 10^{-7})\Rightarrow
p H = − l o g ( 2.5 ) − l o g ( 1 0 − 7 ) ⇒ pH=-log(\sqrt{2.5})-log(10^{-7})\Rightarrow
p H = − 1 2 l o g ( 2.5 ) + 7 ⇒ pH=-\frac{1}{2}log(2.5)+7\Rightarrow
p H ≈ 6.8 \boxed{pH\approx 6.8}

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