PH = 4; [H3O

pH = 8; [H3O

^{+}]= 10^{-4}pH = 8; [H3O

^{+}]= 10^{-8}pH = 12;^{ }[H3O+]=10^{-12}^{}^{ }PH = 4; [H3O^{+}]= 10^{-4}

pH = 8; [H3O^{+}]= 10^{-8}
pH = 12; ^{ }[H3O+]=10^{-12}
^{}

^{ }

pH = 8; [H3O

Ph 11.4 as its near alkaline because of oxygen

Hydroxide concentration i.e. [OH-] of the pH=4,8 &12 can be calculated using the following two formulae: 1) pH = - log [H+] where, [H+] represents the Hydrogen ion concentration. Using this formula, [H+] can be calculated. Which will be for pH=4, [H+] = 1x10^-4 for pH = 8,[H+] = 1x10^-8 and for pH =12, [H+] = 1x10^-12. Now to find the [OH-] for the three solutions, we have to use another formula: 2) For any aqueous solution [H+] x [OH-] = 1x10^-14 (This is known as the ionic product of water and is represented as Kw). Thus for the 1st solution where pH=4, [H+]=1x10-4 so the [OH-] will be = 1x10^-10. Similarly, for pH=8, [OH-]=1x10^-6 and for pH=12, [OH-] = 1x10^-2.

A) 1 x 10 exponent -10

b) 1 x 10 exponent - 6

c) 1 x 10 exponent -2

using your calculator enter formula 10 exponent -pH.

Since your pH isn't decimals simply subtract hydrogen concentration from 1 x 10 exponent -14

b) 1 x 10 exponent - 6

c) 1 x 10 exponent -2

using your calculator enter formula 10 exponent -pH.

Since your pH isn't decimals simply subtract hydrogen concentration from 1 x 10 exponent -14

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