10pKa=10−log(Ka){\displaystyle 10^{pKa}=10^{-log(Ka)}} 10−pKa=10log(Ka){\displaystyle 10^{-pKa}=10^{log(Ka)}} 10−pKa=Ka{\displaystyle 10^{-pKa}=Ka} → Ka=10−pka{\displaystyle Ka=10^{-pka}} For this example, we’ll say pKa=14{\displaystyle pKa=14} (which is in fact the pKa for plain water). [2] X Research source
Type in 10{\displaystyle 10}. Press the xy{\displaystyle x^{y}} button. Type −14{\displaystyle -14} in the superscript box. Press the ={\displaystyle =} button.
1e−14{\displaystyle 1e-14} equals 1∗10−14{\displaystyle 110^{-14}}. So a pKa of 14{\displaystyle 14} equals a Ka of 1∗10−14{\displaystyle 110^{-14}}. In some cases, your calculator may provide the answer in decimal form, such as Ka=0. 00072{\displaystyle Ka=0. 00072} instead of its equivalent Ka=7. 2∗10−4{\displaystyle Ka=7. 2*10^{-4}}. The latter format is typically how Ka is represented, however.
Converting to the -log means that a lower pKa value indicates a stronger acid, as opposed to the Ka, in which a higher Ka value indicates a stronger acid. In this example, let’s make Ka=4∗10−4{\displaystyle Ka=4*10^{-4}} (nitrous acid).
Press the −{\displaystyle -} button. Press the log{\displaystyle log} button. Enter 4∗10{\displaystyle 4*10} inside the parentheses. Press the xy{\displaystyle x^{y}} button. Enter −4{\displaystyle -4} in the superscript box. Press the ={\displaystyle =} button.
Finding an estimate of pKa from Ka is easier. Say, for example, Ka equals 2. 33∗10−11{\displaystyle 2. 3310^{-11}}: Based on this Ka and the “equation” given above, n=2. 33{\displaystyle n=2. 33} and m=11{\displaystyle m=11}. Round n{\displaystyle n} to a single digit if needed (2 in this case), then plug these into the second half of the “equation. ” (11−1). (10−2){\displaystyle (11-1). (10-2)} →10. 8{\displaystyle 10. 8}. Your estimate of Ka is 10. 8, while the actual answer is 10. 63. Estimating Ka from pKa (let’s say it equals 10. 63) is a bit trickier and less accurate: Round to a single digit after the decimal as needed: 10. 63 → 10. 7 Based on the (m−1). (10−n){\displaystyle (m-1). (10-n)} half of the “equation,” add 1 to the 10 (from 10. 7) to get m{\displaystyle m} and subtract the 7 (from 10. 7) from 10 to get n{\displaystyle n}; so m=11{\displaystyle m=11} and n=3{\displaystyle n=3}. Plug the results (m=11 & n=3) into the first half of the “equation”: 3∗10−11{\displaystyle 310^{-11}} is your estimate for Ka, while the actual answer is 2. 33∗10−11{\displaystyle 2. 33*10^{-11}}.
Ka=10−pKa{\displaystyle Ka=10^{-pKa}} Ka=10−3. 14{\displaystyle Ka=10^{-3. 14}} Ka=0. 00072{\displaystyle Ka=0. 00072} → Ka=7. 2∗10−4{\displaystyle Ka=7. 2*10^{-4}}[6] X Research source
Ka=10−pKa{\displaystyle Ka=10^{-pKa}} Ka=10−11. 65{\displaystyle Ka=10^{-11. 65}} Ka=2. 2e−12{\displaystyle Ka=2. 2e^{-12}} → Ka=2. 2∗10−12{\displaystyle Ka=2. 2*10^{-12}}[7] X Research source
pKa=−log(Ka){\displaystyle pKa=-log(Ka)} pKa=−log(2. 33∗10−11){\displaystyle pKa=-log(2. 33*10^{-11})} pKa=10. 63{\displaystyle pKa=10. 63}[8] X Research source
pKa=−log(Ka){\displaystyle pKa=-log(Ka)} pKa=−log(6. 46∗10−5){\displaystyle pKa=-log(6. 46*10^{-5})} pKa=4. 19{\displaystyle pKa=4. 19}[9] X Research source
Using n∗10−m=(m−1). (10−n){\displaystyle n*10^{-m}=(m-1). (10-n)}, plug 5 (from 5. 6) in as n{\displaystyle n} and 6 (from 5. 6) in as m{\displaystyle m} in the second half of the “equation. ” (10-1). (10-6) → 9. 4. The accurate result for PKa is 9. 25. [10] X Research source
