THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE
Acids have pH value less than 7 and bases higher , so NaOH has higher pH value.
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH)7X = 4.692X = 0.7 M HCl==========
Balanced equation. NaOH + HCl -> NaCl + H2O all one to one. find moles HCl. 11 grams HCl (1 mole HCl/36.458 grams ) = 0.3017 moles HCl Moles HCl same as moles NaOH Molarity = moles of solute/Liters of solution 1.06 M NaOH = 0.3017 moles NaOH/liters of solution = 0.2846 Liters this is equal to..... 285 milliliters of NaOH needed
NaOH
.17
HCl+NaOH, when mixed in equimolar amounts, produces a neutral solution of NaCl.
(7 mL)(X M HCl) = (27.6 mL)(0.170 M NaOH)7X = 4.692X = 0.7 M HCl==========
Balanced equation. NaOH + HCl -> NaCl + H2O all one to one. find moles HCl. 11 grams HCl (1 mole HCl/36.458 grams ) = 0.3017 moles HCl Moles HCl same as moles NaOH Molarity = moles of solute/Liters of solution 1.06 M NaOH = 0.3017 moles NaOH/liters of solution = 0.2846 Liters this is equal to..... 285 milliliters of NaOH needed
NaOH
.17
HCl+NaOH, when mixed in equimolar amounts, produces a neutral solution of NaCl.
NaOH
The concentration of hcl is 0.13.
NaOH and HCl
HCl + H_2O NaOH + HCl H_2 + N_2 HNO_3 + NaCl KOH + H_2O
ewan ko sa inyo
Given: 27 mL of NaOH, 0.45M; 20 mL HCI Need: M of HCI 27 ml NaOH*(1 L NaOH/1000mL NaOH)*(0.45M NaOH/1L NaOH)*(1mole HCI/1 mole NaOH)=0.012 0.012/0.02=0.607 M HCI (or rounded 0.61 M HCI)
The product is sodium chloride.The reaction is:NaOH + HCl - NaCl + H2O