Acid-base theory
Buffers
Exercise 4

Calculate the change in $pH$ achieved by adding $5.0\; mL$ $H Cl$ $0.100\; M$ to
a) $95.00 \; mL$ of a $0.100\; M$ $NH_3$ and $0.100\; M$ $NH_4Cl$ solution
$\Delta pH$ $ = $ $0.046 $
b) $95.00 mL$ water
$\Delta pH$ $ = $ $4.70 $