This page looks at calculations resulting from acid-alkali titrations. Don't even think about reading this page unless you are happy with all the other calculations in this section. In particular, it follows on from the page about calculations involving solutions. The page doesn't cover the practical details of doing titrations. You will find these in detail in the second part of the page about making soluble salts. Read the part headed "Making sodium, potassium and ammonium salts" before you carry on with this page.
A titration is used to find an unknown concentration of a solution by reacting it with a solution of known concentration. The examples will make this clear.
25.0 cm Hydrochloric acid of unknown concentration was run in from a burette until the indicator turned from yellow to orange, showing the solution to be neutral. 20.0 cm
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important: In this, and the rest of these calculations, I am expressing small numbers in scientific notation (for example, 2.50 x 10^{-3}) because that is almost certainly how it will come off your calculator.
Your calculator will show this as 2.5 | |

Start with what you know most about. You know the volume and concentration of the sodium hydroxide, so start there and calculate the number of moles. You know that 1000 cm
This is all pretty obvious, apart possibly from the last step. If this worries you, insert another step involving 1 cm
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Note: Whichever way you do this, you can just leave the intermediate answers on your calculator ready for the next step. If you write down the answers step at a time and they need rounding, you could round them to 1 more significant figures than your answer needs - but still go on using the full number on your calculator. | |

This example adds another step because you are going to have to calculate the concentration of one of the substances in g dm 25.0 cm
Calculate the concentration of the sodium carbonate in (RAMs: H = 1; C = 12; O = 16; Na = 23)
This time you know everything about the hydrochloric acid, so start there.
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Note: In this example, it is obvious that if the reaction is 1 : 1, and the sodium hydroxide is twice as concentrated, it will only need half as much as the volume of HCl. So why go to all this bother?
I have deliberately chosen the numbers so that they are easy to work with. That won't always be the case, and the reaction you are looking at won't necessarily be 1 : 1. You need to know how to tackle this whatever reaction or numbers you are given. | |

It is really important tha you practise doing the calculations in the form that your examiners will use. Get hold of as many past papers and mark schemes as you can, and work through all the examples you can find. You will probably find that the way they ask questions is fairly repetitive, and you need to get used to the way they do it. **Where would you like to go now?**-
**To the calculations menu . . .**
© Jim Clark 2021 |