1.0 g KOH is added to the buffer. First change the mass to moles: 1.0 g KOH × 1 mol KOH/(39.1 g + 16.0 g + 1.0 g) = 0.018 mol in 1.00 L
[KOH] = 0.018 M
(The pH of this strong base in the absence of buffer would be pOH = –log(0.018) = 1.74 so pH = 14.00 – 1.74 = 12.26)