Alkalinity is defined as a measure of the buffering capacity of water to neutralize strong acid $($i$.$e$.$ the water's ability to prevent its own pH from changing when an acid is added to the water$).$ This capacity is attributed to bases that are present in natural waters including OH$-,$ HCO$3-,$ and CO$32-.$ More alkalinity in your water sample means more buffering capacity of your water sample.

In order to determine the alkalinity of your samples, samples will be titrated with hydrochloric acid to a certain pH endpoint $($pH $4\mathrm{.}5$ for Total Alkalinity$),$ and the volume of the acid used for the titration will be recorded.

The following reactions are occurring during titration;

pH range above $8\mathrm{.}3$

OH$-+$ H$+->$ H$2$O

CO$32-+$ H$+->$ HCO$3-$

pH range between $8\mathrm{.}3$ and $4\mathrm{.}5$

HCO$3-+$ H$+->$ H$2$CO$3$

If carbonate $($CO$32-)$ is present in your sample, CO$32-$ will consume one H$+$ when the solution is titrated to pH $8\mathrm{.}3,$ and it will consume another H$+$ during further titration from pH $8\mathrm{.}3$ to pH $4\mathrm{.}5.$ All of your samples should have a pH below $8\mathrm{.}3,$ so a certain amount of acid is required to reach pH $4\mathrm{.}5,$ and thus your sample contains only HCO$3-($no OH$-$ and CO$32-).$

Calculation

Total Alkalinity $=$ Amount of acid used to reach pH $4\mathrm{.}5($mL$)*$ molarity of acid $($mol$/$L$)*50,000($mg CaCO$3/$mol$)/$ sample volume $($mL$)$

NOTE: the $50,000($mg CaCO$3/$mol$)$ is a constant.

This can also be written as

For example, a former student performed this experiment and recorded the following data $($NOTE: your values will be different!$)$:

Trial $1$ Trial $2$

Volume of water sample in Erlenmeyer flask $($mL$)50\mathrm{.}0025\mathrm{.}00$

Molarity of acid in buret $($M$)0\mathrm{.}018790\mathrm{.}01948$

Initial volume reading of acid in buret $($mL$)1\mathrm{.}453\mathrm{.}18$

Final volume reading of acid in buret at endpoint $($mL$)3\mathrm{.}184\mathrm{.}10$

Total volume of acid added $($final $-$ initial$)($mL$)1\mathrm{.}73$

total alkalinity $($CaCO$3$ mg$/$L$)$

When they perform their calculation for Trial $1,$ they get the following:

They changed a few things for Trial $2($volume of water sample and molarity of acid$)$ which you will NOT be doing; you'll use a consistent volume of water and acid molarity. Regardless, go ahead and calculate the total alkalinity from the data they got in their second trial. Be sure to calculate the total volume of acid

Answer rating: 100% (QA)

First lets calculate the total volume of acid added in Trial 2 This is done by subtract... View the full answer