Question

$X$ mL of $0.001M$ $HCl$ should be added to $10{cm}^3$ of $0.001N$ $NaOH$ to change its $pH$ by one unit. The nearest integer to X is

Answer 1

A pH change of 1 unit corresponds to pOH change of 1 unit.
pOH of 0.001 N NaOH is 3. It changes to 4. The hydroxide ion concentration changes to 0.0001 N.
Number of millinoles of NaOH initially present $=0.001N\times 10mL=0.01millimoles$.
Number of millimoles of HCl reacted $=0.001N\times 10mL=0.001X$
Number of millimoles of NaOH remaining $=0.01-0.001X$
Total volume $=10+Xml$
The normality of NaOH $=\frac{0.01-0.001X}{10+X}=0.0001N$
Hence, $X=8$
Thus, $8$ ml of $0.001M$ HCl should be added.