# What is the resulting pH if 10 millimoles of HCl is added to 1 liter of a 0.1 M phosphate buffer at pH 7.00 (pK=6.82)?

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### 1 Answer

The phosphate buffer originally has a pH of 7.00

From Henderson- Hasselbalch equation,

`7.00=6.82+log|[HPO4^(2-)]/[ H2PO4^-]|`

`log|[HPO4^(2-)]/[ H2PO4^-]|=0.18`

`|[HPO4^(2-)]/[ H2PO4^-]|=10^(0.18)=1.5136`

Let the molar concentration of `HPO4^(2-)` be x

Therefore, `x/[ H2PO4^-]=1.5136`

`[H2PO4^-]=x/1.5136=0.6607 x`

By condition `x+0.6607x=0.1`

`rArr [HPO4^(2-)] =x=0.1/1.6607=0.060216 M`

and `[H2PO4^-]=x/1.5136=0.039784 M`

On addition of 10 millimoles of HCl, 10 millimoles of ` [HPO4^(2-)] ` would be converted to `[H2PO4^-]`

`[HPO4^(2-)]=0.060216-10*10^(-3)=0.050216 M`

`[H2PO4^-]=0.039784 +10*10^(-3)=0.049784 M`

`pH=6.82+log(0.050216/0.049784)=6.8237`

`~~6.82`