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A 100 ml $\ce{HCl}$ solution has a pH of $3.7$. You want the solution to be of pH 4.5. You have a solution of $10\ \mathrm M$ $\ce{NaOH}$.
Since 1 hydroxide ion is created from each molecule of sodium hydroxide that dissociates , hence 0.02 M NaOH will give 0.02 M of hydroxide ions. p O H = − l o g 10 (O H) − = − l o g 10 (0.02) p O H = 1.70 p H + p O H = 14 ⇒ p H = 14 − 1.70 = 12.3 p H = 12.3
$3\\cdot 10^{-2}\\ \\mathrm{mol}$ of $\\ce{NaOH}$ are added to a $1\\,\\mathrm{l}$ solution of $2\\cdot10^{-2}~\\mathrm{M}\\ \\ce{CH3COOH}$. Find the $\\mathrm{pH ...
You have 100 ml of NaOH @ pH 13. If you dilute by 10 times, then the new pH will be 12, so you'll need to dilute it another 10 times to reach pH 11. If you start with 100 ml, you'll need to end with 10 litres in total (100 folds). You already have your 100 ml, so you'll need to add 10000 ml −100 ml =9900 ml.
$\begingroup$ Do you really need to know pH or just that solution is 4.00 molar NaOH? Can you assume that the solution is "pure" NaOH? No NaCl, no KOH. I'm wondering about a sodium electrode, or conductivity. Carbonate from the air will be a problem. $\endgroup$ –
That is, 5 moles of NaOH N a O H gives 5 moles of OHX− O H X − ions after complete dissociation. Find the equivalents of HX+ H X + from phosphoric acid (a) and the equivalents of OHX− O H X − from NaOH N a O H (b). Also calculate the total volume. Number of equivalents= NV N V. But be careful since concentration is given in molarity.
A disodium salt of EDTA also has two remaining carboxylic-acid groups, which can be deprotonated by addition of base, like $\ce{NaOH}$. By the way, higher pH is generally better especially if you plan to use it for chelation, i.e. cation-capture. The labs I've taught in the past use buffers of pH ~10 with EDTA for complexometric titration ...
So I wanted to know what the reaction between sodium hydroxide and carbon dioxide can be, and upon research I got 2 answers. The first one is. CO. and the second one is a two-step reaction, first with water then sodium hydroxide: COX2 + HX2O(l) HX2COX3(aq), HX2COX3(aq) + NaOH(aq) NaHCOX3(aq) + HX2O(l) So my question is which one of these ...
2. I have a buffer containing 0.2 M of the acid HA H A, and 0.15 M of its conjugate base AX− A X −, with a pH of 3.35. I need to find the pH after 0.0015 mol of NaOH N a O H is added to 0.5 L of the solution. I started by converting everything to moles, writing out an equation for HA H A and OHX− O H X − reacting to make HX2O H X 2 O ...
I was under the impression that the reaction for such a high pH solution would look like this: 2NaOH(aq) + COX2 − ⇀ ↽ − NaX2COX3(aq) + HX2O. But the results I have got don't fit this. For example the rate of change of [NaOH] and [NaX2COX3] should be a 2:1 ratio but I got 0.849:1. This is a significant difference.