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| pH = - log10 [H3O+] | Doing this on calculator | displays this |
| Example 1: Calculate the pH of a solution where [H3O+] is 6.02 × 10-4 mol L–1. Answer: pH = - log10(6.02 × 10–4) pH = – (–3.22) = 3.22 | Enter digits before × then E (or EXP) | 6.02e |
| then the exponent | 6.02e-4 | |
| then log | -3.22 | |
| then change sign (+/-) | 3.22 |
| [H3O+] = 10–pH | Doing this on calculator | displays this |
| Example 2: Calculate [H3O+] for a solution where pH 4.76. Answer: -log10[H3O+] = 4.76 log10[H3O+] = -4.76 [H3O+] = 10–4.76 [H3O+] =1.74 ×10–5 | numbers then +/- | –4.76 |
| SHFT(2nd) then log or 10^ | 10^( ) | |
| Enter or = | 1.7378–05 | |
| Write the result as | 1.74×10–5 |