The pH of a 0.001M NaOH will be
3
2
11
12
0.001 M of NaOH means \([OH^-]\) = 0.001 = 10−3M ⇒ pOH = 3 pH + pOH = 14 ⇒ pH = 14 – 3 = 11
The magnitude of the projection of 2^i+3^j+4^k on the vector ^i+^j+^k will be –––––