# A 40.0 mL volume of 0.100 M HCl is titrated with 0.100 M NaOH [Solved]

= 4- 1.2304
= 2.769
so pH = 14- pOH = 11.23 [ pH pOH = 14 ]

complete quantity = 40 45 = 85 ml
40 ml of NaOH will probably be neutralised by HCl.
remaining= 5 ml 0.1 M NaOH in 85ml
[OH- ] = 0.1x 85/5000 = 0.0017
so pOH = log[OH-]
= log[ 0.0017]

M1V1 = M2V2 (0.0100 M)(V1)(2) = (0.0150 M)(0.250 L) V1 = amount of HCl = 0.750 L H2CO3 is diprotic; it notably is how I acquired right here up with these steps whereas exhibiting my work. i am not 100% specific, but a minimal of answering the query is further helpful than not answering. =)

= 4- 1.2304
= 2.769
so pH = 14- pOH = 11.23 [ pH pOH = 14 ]

M1V1 = M2V2 (0.0100 M)(V1)(2) = (0.0150 M)(0.250 L) V1 = amount of HCl = 0.750 L H2CO3 is diprotic; it notably is how I acquired right here up with these steps whereas exhibiting my work. i am not 100% specific, but a minimal of answering the query is further helpful than not answering. =)

Extra base = 5 mL
Moles of NaOH within the extra base = (5 mL)*( 0.100 M NaOH) = 0.5 millimoles of NaOH
(we might convert this to moles however we do not have to, as will probably be seen shortly)
Complete quantity = 40 mL (of HCL) 45 mL (of NaOH) = 85 mL
Focus of NaOH in complete resolution= (0.5 millimoles of NaOH)/(85 mL) = 0.00588 M
(1 mmol/mL = 1 mol/L)
pOH = -log[OH-}
pOH = -log (0.00588 M)
pOH = 2.23
pH pOH = 14
pH = 14 pOH
pH = 14 2.23
pH = 11.77

This post is last updated on hrtanswers.com at Date : 1st of September – 2022